Download Unit 9 Factoring - the Algebra Class E

Unit 9: Factoring
Lesson 1: Introduction to Factoring
Factors and Multiples
6 ● 5 = 30
6 ● 8 = 48
_________
_______
___ and ____ are _________ of _____.
___ and ____ are _________ of _____.
_________ is a multiple of ________.
_________ is a multiple of ________.
_________ is a multiple of ________.
_________ is a multiple of ________.
Common Factors
If two numbers are divisible by the same number, then they are called common factors
___ is a common factor of ______ and ________.
Try These on Your Own
1. List the factors of 72.
2. 36 is divisible by _____.
3. _____ is a multiple of 12.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
4. ______ is a common factor of 32 & 40.
Unit 9: Factoring
Prime and Composite Numbers
Prime Numbers
Composite Numbers
Greatest Common Factor
The Greatest Common Factor is the greatest integer that is a common factor
of two or more numbers.
Example 1
Example 2
What is the greatest common factor for 18
and 27?
What is the greatest common factor for
18
27
3x and 9x2
9x2
3x
Example 3
Example 4
What is the greatest common factor of
What is the greatest common factor of
6x2 y and 9x3 y2 ?
2x and 3y?
6x2 y
9x3 y2
2x
3y
If: 2x + 3y
This polynomial is ____________ because
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 1: Introduction to Factoring Practice
Part 1: Read each question and answer completely.
1. 35 is divisible by _______ (give 2 solutions)
2. Name 6 factors of 12.
3. ________ is a multiple of 13. (Give 3 solutions)
4. _______ is a common factor of 36 and 48.
5. ______ is a common factor of 144 and 120.
6. ______ is the greatest common factor of 48 and 64.
7. ______ is the greatest common factor of 48 and 60.
8. Is 9 a prime number? Explain why or why not.
9. 15 is a composite number. Explain why.
10. Do you think that the number 1 is prime, composite, or neither? Explain.
Part 2: Find the greatest common factor (GCF) for each set of numbers.
1. 12a2 and 14 a
7. 3xy,
2x2y,
2. x2 y3 and 2xy2
8. 3a2b, 27ab2, and 9ab
3. 3ab5 and 4a3b3
9. 2x2y3z,
4. 9rs and 12r2s
10. 15rs3t, 7r2st, and 5r2s2t
5. 4a3b2 and 2a3
11. 18xz, 9x2y4z2, and 3xyz
6. 16x4yz2 and 24x3z
12. 7x2y5, 5x5y4, and 8x3y5
and
4x3y
6xy2z2, and 18xz3
Find the greatest common factor (GCF) for each set of numbers.
(2 points each)
1. 13x3yz and 26xy2z2
2. 18r5s8t4, 81r6s5t6, and 27r9s7t5
3. 22abc and 11a2b
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 1: Introduction to Factoring Practice – Answer Key
Part 1: Read each question and answer completely.
1. 35 is divisible by 7 and 5 (give 2 solutions)
2. Name 6 factors of 12.
(7 & 5 are factors of 35)
(1,2,3,4,6,12)
(1 x12) (2 x6) (3x4)
3. 26, 39, 52 is a multiple of 13. (Give 3 solutions) (13 x2 = 26) (13x3 = 39) (13 x4 = 52) (You may
also have had: 65, 78, 91, 104, 117, 130 …)
4. 6 is a common factor of 36 and 48. (6 x6 = 36) (6 x 8 = 48) (You may also have had 4 because
(4 x9 = 36) (4 x 12 = 48). Other answers include: 2 , 3, and 12
5. 12 is a common factor of 144 and 120. (12 x12 = 144) (12 x 10 = 120) (Other answers include:
2, 3, 4, 6, 8, 24)
6. 16 is the greatest common factor of 48 and 64.
Factors of 48:
Factors of 64:
1, 2, 3, 4,6, 8, 12, 16, 24 ,48
1, 2, 4, 8, 16, 32, 64
The common factors are: 1,2,4, 8, 16
The greatest common factor would be 16.
7. 12 is the greatest common factor of 48 and 60.
Factors of 48:
Factors of 60:
1, 2, 3, 4,6, 8, 12, 16, 24 ,48
1,2,3,4,5,6,10,12,15,20,30,60
The common factors are: 1,2,3,4,6, 12
The greatest common factor would be 12.
8. Is 9 a prime number? Explain why or why not.
9 is not a prime number. A prime number only has factors of 1 and itself. 9 has the following factors:
1,3,9. Since 3x3=9, 9 is not a prime number.
9. 15 is a composite number. Explain why.
15 is a composite number. A composite number has one or more factors other than itself and 1. (It’s the
opposite of a prime number.)
The factors of 15 are: 1,3,5,15. Since 3x5=15, we can classify 15 as a composite number.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
10. Do you think that the number 1 is prime, composite, or neither? Explain.
Since the only factor of 1 is 1, it is not considered to be prime or composite.
The prime numbers less than 50 are:
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47.
Part 2: Find the greatest common factor (GCF) for each set of numbers.
1. 12a2 and 14 a
GCF: 2a
The GCF for 12 and 14 is 2.
The GCF for a2 and a is a
2. x2 y3 and 2xy2
7. 3xy,
2x2y,
and
4x3y
GCF: xy
The GCF for 3,2, and 4 is 1. (We don’t
need to write)
The GCF for x, x2 and x3 is x.
The GCF for y, y, and y is y
8. 3a2b, 27ab2, and 9ab
GCF: xy2
GCF: 3ab
The GCF for 1 and 2 is 1. (We don’t
need to write.)
The GCF for x2 and x is x.
The GCF for y3 and y2 is y2
The GCF for 3, 27, and 9 is 3.
The GCF for a2 and a is a.
The GCF for b, and b2 is b.
3. 3ab5 and 4a3b3
9. 2x2y3z,
6xy2z2, and 18xz3
GCF: ab3
GCF: 2xz
The GCF for 3 and 4 is 1. (We don’t
need to write.)
The GCF for a3 and a is a.
The GCF for b5 and b3 is b3.
The GCF for 2, 6 and 18 is 2.
The GCF for x2 and x is x.
Since 18xz3 does not have a y term, we
cannot include a y in our GCF.
The GCF for z3, z2 and z is z.
4. 9rs and 12r2s
10. 15rs3t, 7r2st, and 5r2s2t
GCF: 3rs
GCF: rst
The GCF for 9 and 12 is 3.
The GCF for r2 and r is r.
The GCF for s is s.
The GCF for 15, 7 and 5 is 1. (We don’t
need to write.)
The GCF for r2 and r is r.
The GCF for s3,s2, and s is s.
The GCF for t is t.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
5. 4a3b2 and 2a3
11. 18xz, 9x2y4z2, and 3xyz
GCF: 2a3
GCF: 3xz
The GCF for 4 and 2 is 2.
The GCF for a3 is a3.
Since 2a3 does not have a b term, we
cannot include a b term in the GCF.
The GCF for 18, 9 and 3 is 3.
The GCF for x2 and x is x.
Since 18xz does not have a y term, we
cannot include a y term in the GCF.
The GCF for z2 and z is z.
6. 16x4yz2 and 24x3z
12. 7x2y5, 5x5y4, and 8x3y5
GCF: 8x3z
GCF: x2y4
The GCF for 16 and 24 is 8.
The GCF for x4 and x3 is x3.
Since 24x3z does not have a y term, we
cannot include a y term in our GCF.
The GCF for z2 and z is z.
The GCF for 7 and 5, and 8 is 1. (We
don’t need to write.)
The GCF for x2, x3, and x5 is x2.
The GCF for y4 and y5 is y4.
1. 13x3yz and 26xy2z2
2. 18r5s8t4, 81r6s5t6, and 27r9s7t5
GCF: 13xyz
GCF: 9r5s5t4
The GCF for 13 and 26 is 13.
The GCF for x3 and x is x.
The GCF for y2 and y is y.
The GCF for z and z2 is z.
The GCF for 18, 81 and 27 is 9.
The GCF for r5, r6 and r9 is r5.
The GCF for s8, s5 and s7 is s5.
The GCF for t4, t6, and t5 is t4.
3. 22abc and 11a2b
GCF: 11ab
The GCF for 22 and 11 is 11.
The GCF for a2 and a is a.
The GCF for b is b.
Since 11a2b does not have a c term, we
cannot include a c term in our GCF.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 2: Factoring Polynomials Using the Greatest Common Factor (GCF)
The Greatest Common Factor (GCF) is the greatest integer that is a common factor
of two or more numbers.
Example 1
Factor: 3x2 + 27x3
Example 2
Factor: 2x2y2 - 8xy2
NOTE: Factoring can always be checked using multiplication!
Example 3
Factor: 3x2y2 + 2x2 + xy2
Example 4
Factoring can make it easier to simplify some
fractions:
Simplify:
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 2: Factoring Polynomials by Using the GCF Practice
Part 1: The greatest common factor has been factored out ( in bold). Fill in the blanks to
complete the factorization of each polynomial.
1. 3x2 + 12xy2 = 3x(______ + ________)
2. 2r2s3 + 5rs6 = rs3(________+_________)
3. 8a2bc4 - 6ab4c4 = 2abc4(___________-___________)
4. 2x2y5 + 4xy3 + 6x2y7 =
2xy3(___________+ ____________+ ____________)
5. 3r3s3t5 – 2r2s7t4 + s5t3 =
s3t3( ___________ - ___________+______________)
Part 2: Factor Completely.
1. 7a2b5 + 21ab4
6. 9x2y – 27xy3 + 3xy
2. 81x2yz – 27xy2z3
7. 12a4b5c + 2a2c + 8a2bc
3. 3a2b3c + 12a3b2c2
8. 22g2h – 11gh2 + g2h2
4. 5s5t2 – 25s4
9. 2w5xy – 3w2x2y + 2w2x
5. 14x4yz2 + 2x4y3z – 7x3yz
10. 8s2t5u2 – 12st2u2 + 2st4u
Part 3: Simplify each fraction by simplifying the numerator.
1.
2.
Factor completely. ( 2 points each)
1. 4x3y5z – 6x3y5
3.
Simplify the fraction. 2. 8r3st + 24r4s2t2
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(2 points)
Unit 9: Factoring
Lesson 2: Factoring Polynomials by Using the GCF Practice- Answers
Part 1: The greatest common factor has been factored out ( in bold). Fill in the blanks to
complete the factorization of each polynomial.
1. 3x2 + 12xy2 = 3x(
x
4y2)
+
2. 2r2s3 + 5rs6 = rs3( 2r + 5s3)
3. 8a2bc4 - 6ab4c4 = 2abc4(
4. 2x2y5 + 4xy3 + 6x2y7 =
4a - 3b3)
2xy3(
5. 3r3s3t5 – 2r2s7t4 + s5t3 =
s3t3(
xy2+ 2+ 3xy4 )
3r3t2 -
2r2s4t +
s2)
Part 2: Factor Completely.
1. 7a2b5 + 21ab4
6. 9x2y – 27xy3 + 3xy
Identify the GCF: 7ab4
Identify the GCF: 3xy
7ab4(ab + 3)
3xy(3x – 9y2 + 1)
2. 81x2yz – 27xy2z3
7. 12a4b5c + 2a2c + 8a2bc
Identify the GCF: 27xyz
Identify the GCF: 2a2c
27xyz(3x – yz2)
2a2c(6a2b5 + 1 + 4b)
3. 3a2b3c + 12a3b2c2
8. 22g2h – 11gh2 + g2h2
Identify the GCF: 3a2b2c
Identify the GCF: gh
3a2b2c(b + 4ac)
gh(22g – 11h + gh)
4. 5s5t2 – 25s4
9. 2w5xy – 3w2x2y + 2w2x
Identify the GCF: 5s4
Identify the GCF: w2x
5s4(st2 – 5)
w2x(2w3y – 3xy + 2)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
5. 14x4yz2 + 2x4y3z – 7x3yz
10. 8s2t5u2 – 12st2u2 + 2st4u
Identify the GCF: x3yz
Identify the GCF: 2st2u
x3yz(14xz + 2xy2 – 7)
2st2u(4st3u – 6u + t2)
Part 3: Simplify each fraction by simplifying the numerator.
1.
2.
Identify the GCF of the numerator: 2xy
2(2 + )
= 2 + 2
Identify the GCF of the numerator: b3
( + 1 − − )
= − − + 1
1. 4x3y5z – 6x3y5
2. 8r3st + 24r4s2t2
Identify the GCF: 2x3y5
Identify the GCF: 8r3st
2x3y5(2z – 3)
8r3st(1 + 3rst)
3.Simplify the fraction. (2 points)
Identify the GCF of the numerator: ab2
(6 + 5 ")
= 6 + 5 "
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 3: Factoring Trinomials
What is a Trinomial?
A trinomial is a polynomial that has _____ terms.
x2 –x - 12
Before we start, let’s review multiplication of polynomials.
Multiply: (x+2)(x+3) =
Multiply: (x – 5) (x+6)
When factoring trinomials in the form of: x2 + bx + c, you must find two numbers whose product
is c and whose sum is b. NOTE: (Always take the sign in front of b and c when determining your
numbers.
Example 1
Example 2
Factor: x2 + 6x + 8
Factor: x2 -9x + 20
2#’s: Product = ____
(
) (
Sum = ____
2#’s: Product = ____
)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Sum = ____
Unit 9: Factoring
Example 3
Example 4
Factor: x2 – 5x – 24
2#’s: Product = _____
Let’s first multiply:
Sum = _______
(x – 3) (x + 3)
Factor: x2 - 81
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 3: Factoring Trinomials Practice
Part 1: Factor by filling in the missing parts.
1. x2 + 8x + 12
Product: _____
Sum: ______
Product: _____
Sum: ______
Product: _____
Sum: ______
Product: _____
Sum: ______
(x + ____) (x + _____)
2. x2 -7x + 12
(x + ____) (x + _____)
3. x2 -x - 12
(x + ____) (x + _____)
2
4. y + y - 6
(y + ____) (y + _____)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Part 2: Factor completely.
1. x2 + 15x + 36
5. x2 - 49
2. x2 + 2x – 24
6. x2 - 36
3. x2 – 14x + 48
7. x2 + 14x + 49
4. x2 – 7x – 30
8. x2 – 17x + 72
Part 3: Thinking Questions
1. What polynomial, when factored, gives (x + 5)(x -6)?
2. What polynomial when factored gives (x – 8) (x+8)?
3. One factor of the trinomial: x2 + 85x + 750 is (x+10), what is the other factor?
4. One factor of the trinomial: x2 – 10 x – 375 is (x – 25), what is the other factor?
5. Explain why the trinomial: x2 – 7x + 14 cannot be factored.
Factor Completely. (2 points each)
1. x2 + 16x + 55
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
2. x2 + 8x + 11
Unit 9: Factoring
Lesson 3: Factoring Trinomials Practice
Part 1: Factor by filling in the missing parts.
1. x2 + 8x + 12
(x + 2 ) (x +
Product: 12
6 )
1 ● 12
1 + 12 = 13
2●6
2+6=8
4●3
4+3=7
Product: 12
2. x2 -7x + 12
(x + -4) (x + -3)
Most commonly written as:
(x-4) (x-3)
(x +
-4) (x +
3)
Most commonly written as:
X
X
Sum: -7
You know that both factors must be negative since
the sum is negative and the product is positive.
-4 ● -3
-4 + (-3) = -7
-6 ● -2
-6 + -2 = -8
-1 ● -12
-1 + -12 = -13 X
Product: -12
3. x2 -x - 12
Sum: 8
X
Sum: -1
Since the product is negative, you know that one
factor must be positive and one must be negative.
Since the sum is negative also, the larger integer will
be negative.
-12 ● 1
-12 + 1 = -11
X
-6 ● 2
-6+ 2 = -4
X
-4 ● 3
-4 + 3 = -1
(x-4) (x+3)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Product: -6
Sum: 1
Since the product is negative, we know that one
integer must be positive and one must be
negative. Also, since the sum is positive, the
larger number must be positive.
4. y2+ y - 6
(y + 3 ) (y + -2)
6 ● (-1)
6 + (-1) = 5
3 ● (-2)
3 + (-2) = 1
X
Most commonly written as:
(y+3)(y-2)
Part 2: Factor completely.
1. x2 + 15x + 36
5. x2 – 49
Need a Sum of 15 and Product of 36.
12 ●3 = 36
and
12+ 3 = 15
Therefore, we can use: 12 and 3
(x+12)(x+3)
Need a Sum of 0 and Product of -49. (There is no
x term, therefore, the sum is 0)
The product is -49, so one integer must be
positive and one must be negative. Since the sum
is 0, the integer should be the same.
7 ● -7 = 49
7 + (-7) = 0
Therefore, we can use: 7 and -7
(x+7)(x-7) This is a Difference of Two Squares
2. x2 + 2x – 24
6. x2 – 36
Need a Sum of 2 and Product of -24.
The product is negative, so one integer must be
positive and one must be negative. Since the sum
is positive, the largest integer should be positive.
6 ● (-4) = -24
and 6 + (-4) = 2
Need a Sum of 0 and Product of -36. (There is no
x term, therefore, the sum is 0)
The product is negative, so one integer must be
positive and one must be negative. Since the sum
is 0, the integer should be the same.
6 ● (-6) = -36 and 6 +(-6) = 0
Therefore, we can use: 6 and -4
(x+6)(x-4)
Therefore, we can use: 6 and -6
(x+6)(x-6) This is a Difference of Two Squares
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
3. x2 – 14x + 48
7. x2 + 14x + 49
Need a Sum of -14 and Product of 48.
Need a Sum of 14 and Product of 49.
The product is positive and the sum is negative.
Therefore, both integers must be negative.
Since both the sum and product are positive, both
integers will be positive.
-6 ● (-8) = 48
and (-6) + (-8) = -14
Therefore, we can use: -6 and -8
7 ●7 = 49 and
7 + 7 = 14
Therefore, we can use: 7 and 7
(x+7)(x+ 7) or (x+7)2
(x-6)(x-8)
This is a perfect square binomial.
4. x2 – 7x – 30
8. x2 – 17x + 72
Need a Sum of -7 and Product of -30.
Need a Sum of -17 and Product of 72.
The product is negative, so one integer must be
positive and one must be negative. Since the sum
is negative, the largest integer should be negative.
Since the product is positive and the sum is
negative, both integers must be negative.
-10 ● 3 = -30 and -10 + 3 = -7
-9 ● (-8) = 72
and -9 + (-8) = -17
Therefore, we can use: -9 and -8
Therefore, we can use: -10 and 3
(x-9)(x-8)
(x+3)(x-10)
Part 3: Thinking Questions
1. What polynomial, when factored, gives (x + 5)(x -6)?
In order to find the original polynomial, we can multiply the two binomials. Use FOIL.
(x+5) (x-6)
x(x) + x(-6) + 5(x) + 5(-6)
x2 – 6x + 5x - 30
x2 – x – 30 is the original trinomial that factors into (x+5) (x-6)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
2. What polynomial when factored gives (x – 8) (x+8)?
This set of binomials is a Difference of Perfect Squares. Notice that they are opposites.
Therefore, the original trinomial is: x2 – 64.
3. One factor of the trinomial: x2 + 85x + 750 is (x+10), what is the other factor?
We are given the trinomial and one of the factors. We know that we must find two numbers whose product
is 750 and sum is 85. One of the numbers is 10. So, 10 times what is 750? 10 ● 75 = 750 AND 10 + 75
= 85. Therefore, the other factor is: (x+75)
4. One factor of the trinomial: x2 – 10 x – 375 is (x – 25), what is the other factor?
We are given the trinomial and one of the factors. We know that we must find two numbers whose product
is -375 and sum is -10. One of the numbers is -25. So, -25 times what is -375? Or, -375/-25 = 15
So, -25 ● 15 = -375 and -25 + 15 = 10 Therefore, the other factor is: (x+15)
5. Explain why the trinomial: x2 – 7x + 14 cannot be factored.
We are looking for two integers that have a product of 14 and a sum of -7. The two integers must both be
negative since the product is positive and the sum is negative. Let’s try:
-1 ● -14 = 14 but -1 + -14 = -15;
-7 ● -2 = 14 but -7 + -2 = -9 There are no other factors of 14 to try.
Therefore, this trinomial cannot be factored.
Factor Completely. (2 points each)
1. x2 + 16x + 55
2. x2 + 8x + 11
Need a Sum of 16 and Product of 55
Need a Sum of 8 and Product of 11.
Since both the sum and product are positive, then
both integers must be positive.
Since both the sum and product are positive, then
both integers must be positive.
11 ● 5 = 55 and 11 + 5 = 16
Therefore, we can use: 11 and 5
11 is a prime integer. Only 1 ● 11 = 11.
Since 1 + 11 ≠ 8, we can conclude that this
trinomial is prime and cannot be factored.
(x+ 11)(x+ 5)
There is no way to factor this trinomial.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Factoring – Quiz # 1
Part 1: Complete each problem. Express your answer in simplest terms. (Use the GCF)
1. Factor the following polynomial using the GCF.
3x2 – 18x + 33
2. Factor the following polynomial using the GCF.
6a5 – 12a4 + 15a3
3. Factor the following polynomial using the GCF.
8x2y3z + 4xy4 z3 – 12x4y5z2
4. Simplify the following fraction:
5. Simplify the following fraction:
6x2y7z5 + 3x2y3z6 + 2xy3 z8
xy3z5
Part 2: Factor each polynomial.
1. x2 + 9x + 18
2. x2 – x – 20
3. y2 – 9y + 14
4. r2 + 2r – 48
5. z2 – 6z + 5
Part 3: Short Answer.
1. Explain the steps used to factor the following polynomial: 3x4 + 6x2 – 3x
2. Is the following polynomial factorable? Explain why or why not.
x2 + 5x + 5
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Unit 9: Factoring
Factoring Quiz #1 – Answer Key
Part 1: Complete each problem. Express your answer in simplest terms. (1 point each)
3x2 – 18x + 33
1. Factor the following polynomial using the GCF.
Step 1: Identify the GCF for 3x2, 18x and 33.
The GCF is 3. Factor a 3 out of each term:
3(x2 – 6x + 11)
Final Answer: 3(x2 – 6x + 11)
6a5 – 12a4 + 15a3
2. Factor the following polynomial using the GCF.
Step 1: Identify the GCF for 6a5, 12a4 and 16a3. The GCF is 3a3. Factor 3a3 out of each term.
3a3(2a2 – 4a + 5)
Final Answer: 3a3(2a2 – 4a + 5)
8x2y3z + 4xy4 z3 – 12x4y5z2
3. Factor the following polynomial using the GCF.
2 3
4
3
4 5 2
Step 1: Identify the GCF for 8x y z, 4xy z , and 12x y z . The GCF is 4xy3z. Factor 4xy3z out of
each term.
4xy3z(2x+yz2 – 3x3y2z)
Final Answer: 4xy3z(2x+yz2 – 3x3y2z)
4.
Simplify the following fraction:
Step 1: Identify the GCF for the numerator: The GCF is: 2a2b3. Factor this out of each term:
2a2b3 (4+ 3ab)
2a2b3
Step 2: Simplify:
2a2b3 (4+ 3ab)
2a2b3
= 4+3ab
Final Answer: 4+3ab
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
5. Simplify the following fraction:
6x2y7z5 + 3x2y3z6 + 2xy3 z8
xy3z5
Step 1: Identify the GCF for the numerator: The GCF is: xy3z5 Factor this out of each term:
xy3z5 (6xy4+3xz+2z3)
xy3z5
Step 2: Simplify:
xy3z5 (6xy4+3xz+2z3)
xy3z5
Final Answer: 6xy4 + 3xz + 2z3
Part 2: Factor each polynomial. (1 point each)
1. x2 + 9x + 18
We need two factors whose product is 18 and sum is 9.
6 & 3 6●3 = 18; 6+3 = 9
Final Answer: (x+6)(x+3)
2. x2 – x – 20
We need two factors whose product is -20 and sum is -1. (Since the product is negative, we need a
positive number and a negative number. The larger number should be negative since the sum is
negative.
4 & -5 4●(-5) = -20; 4+(-5) = -1
Final Answer: (x+4)(x-5)
3. y2 – 9y + 14
We need two factors whose product is 14 and sum is -9. (Since the product is positive and the sum is
negative, we must have two numbers that are negative.
-7& -2 -7●(-2) = 14; -7+(-2) = - 9
Final Answer: (y-7)(y-2)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
4. r2 + 2r – 48
We need two factors whose product is -48 and sum is 2. (Since the product is negative, we need a
positive and a negative number. The larger number will be positive since the sum is positive.
-6 & 8 -6●8 = -48; -6+8 = 2
Final Answer: (r-6)(r+8)
5. z2 – 6z + 5
We need two factors whose product is 5 and sum is -6. (Since the product is positive and the sum is
negative, we must have two number that are negative.)
-5 & -1
-5●(-1) = 5;
-5+(-1) = -6
Final Answer: (z-5)(z-1)
Part 3: Short Answer. (2 points each)
1. Explain the steps used to factor the following polynomial: 3x4 + 6x2 – 3x
First you must look to see if there is a greatest common factor. The GCF is 3x
Then factor out the 3x.
3x(x3 + 2x – 1) This is the final answer.
***Look for key words: greatest common factor***
2. Is the following polynomial factorable? Explain why or why not.
x2 + 5x + 5
This polynomial is NOT factorable. We would need to find two positive numbers whose product is 5
and whose sum is 5. The only factors of 5 are 5 and 1. Since 5+1 = 6 and not 5, there is no way to
factor this polynomial using integers.
This quiz is worth 14 points.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 4: Factoring Trinomials: A Special Situation
Using two different factoring methods.
In the previous lesson, we factored trinomials that had a lead coefficient of 1. Of course,
many trinomials have a lead coefficient that is greater than 1. We will explore one method of
factoring these types of trinomials.
Example 1
Factor: 3x2 – 3x – 36
Notice how the lead coefficient is not 1. However, each term is divisible by 3.
Step 1:
Note: If a trinomial looks difficult to factor, first check to see if a
Greatest Common Factor (GCF) can be factored out.
Example 2
Factor: 4x3 + 4x2 – 24x
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 4: Factoring Trinomials (A Special Situation) Practice
Part 1: Factor Completely
1. 2x2 + 10x – 48
7. 2y5 – 16y4 + 32y3
2. 5y2 – 40y + 60
8. 3x2 - 243
3. 6x3 + 54x2 + 120x
9. 2b3 – 2b2 – 24b
4. 3s3 – 108s
10. 4x3 – 20x2 + 24x
5. 5x4 – 50x3 + 125x2
11. 6w4 – 18w3 – 24w2
6. 2a4 + 12a3 + 18a2
12. 3x2 – 9x - 54
Part 2: Thinking Questions
1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored.
2. Give at least one example of a trinomial without a lead coefficient of one that can be factored.
3. Give an example of a prime trinomial.
Factor each trinomial. (3 points each)
1. 3x3 +18x2 – 81x
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
2. 2x3 – 8x2 – 8x
Unit 9: Factoring
Lesson 4: Factoring Trinomials (A Special Situation) Answers
Part 1: Factor Completely
1. 2x2 + 10x – 48
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
2 is the GCF for all terms.
2(x2 + 5x – 24)
Step 2: Factor the expression inside of
parenthesis.
We need a positive and negative number that
have a product of -24 and a sum of 5
(8 & 3, 8 must be positive since we need
a +5 for a sum).
Final Answer: 2(x+ 8) (x-3)
7. 2y5 – 16y4 + 32y3
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
3
2y is the GCF for all terms.
2y3(y2 – 8y + 16)
Step 2: Factor the expression inside of
parenthesis.
We need 2 negative numbers that have a product
of 16 and a sum of -8.
(-4 & -4).
2y3(y-4) (y-4)
Final Answer: 2y3(y-4)2
2. 5y2 – 40y + 60
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
5 is the GCF for all terms.
5(y2 - 8x + 12)
Step 2: Factor the expression inside of
parenthesis.
We need 2 negative numbers that have a product
of 12 and a sum of -8
(-6 & -2).
8. 3x2 – 243
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
3 is the GCF for all terms.
3(x2 - 81)
Step 2: Factor the expression inside of
parenthesis.
This is a difference of two squares. So,
what number squared is 81? (9)
Final Answer: 3(x+ 9) (x-9)
Final Answer: 5(y-6) (y-2)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
3. 6x3 + 54x2 + 120x
9. 2b3 – 2b2 – 24b
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
6x is the GCF for all terms.
6x(x2 + 9x + 20)
Step 2: Factor the expression inside of
parenthesis.
We need 2 positive numbers that have a product
of 20 and a sum of 9
(4 & 5)
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
2b is the GCF for all terms.
2b(b2 - b – 12)
Step 2: Factor the expression inside of
parenthesis.
We need a positive and negative number that
have a product of -12 and a sum of -1 (4 & 3, 4 must be negative since we need
a -1 for a sum).
Final Answer: 6x(x+ 4) (x+ 5)
Final Answer: 2b(b-4) (b+3)
4. 3s3 – 108s
10. 4x3 – 20x2 + 24x
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
3s is the GCF for all terms.
3s(s2 - 36)
Step 2: Factor the expression inside of
parenthesis.
This is a difference of two squares. What
number when squared equals 36? (6)
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
4x is the GCF for all terms.
4x(x2 – 5x+ 6)
Step 2: Factor the expression inside of
parenthesis.
We need two negative numbers that have
a product of 6 and a sum of -5. (-3 & -2)
Final Answer: 3s(s+ 6) (s-6)
Final Answer: 4x(x-3) (x-2)
5. 5x4 – 50x3 + 125x2
11. 6w4 – 18w3 – 24w2
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
2
5x is the GCF for all terms.
5x2(x2 -10x + 25)
Step 2: Factor the expression inside of
parenthesis.
We need two negative numbers whose
product is 25 and sum is -10 (-5 & -5)
This is a perfect square.
Final Answer: 5x2(x - 5) (x-5) or
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
2
6w is the GCF for all terms.
6w2(w2 -3w -4)
Step 2: Factor the expression inside of
parenthesis.
We need a positive and negative number
whose product is -4 and whose sum is -3.
(-4 & 1) 4 must be negative since the
sum is negative.
5x2(x-5)2
Final Answer: 6w2(w -4) (w +1)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
6. 2a4 + 12a3 + 18a2
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
2
2a is the GCF for all terms.
2a2(a2 + 6a + 9)
Step 2: Factor the expression inside of
parenthesis.
This is a perfect square. What number
can you square and get 9, and add to get
6? (3)
12. 3x2 – 9x – 54
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
3 is the GCF for all terms.
3(x2 – 3x - 18)
Step 2: Factor the expression inside of
parenthesis.
What positive and negative numbers have
a product of -18 and a sum of -3? (-6 & 3)
6 must be negative since the sum is
negative.
Final Answer: 2a2(a+3)2
Final Answer: 3(x -6) (x +3)
Part 2: Thinking Questions
1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored.
These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards
and start with two binomials. You can choose any binomials.
For example, I’ll choose (x+4) (x+2) now multiply using foil.
(x)(x) +(2)x+(4)x+4(2)
x2 +6x + 8 is a trinomial that can be factored. We know this because we started with the two factors and
multiplied.
2. Give at least one example of a trinomial without a lead coefficient of one that can be factored.
These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards
and start with two binomials. You can choose any binomials. Since you want a lead coefficient that is
greater than 1, make sure that one of your binomials has a coefficient of x that is greater than 1.
For example, I’ll choose (2x+4) (x+2) now multiply using foil.
(2x)(x) +(2x)2+(4)x+4(2)
2x2 +8x + 8 is a trinomial that can be factored. We know this because we started with the two factors and
multiplied.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
3. Give an example of a prime trinomial.
A prime trinomial cannot be factored. For example:
x2 + 2x + 2 - We would need two positive numbers whose product is 2 and whose sum is 2. The only way
to make a product of 2 is (2 ●1) Since 2+1 = 3, we know that this cannot be factored.
The best way to make up a prime trinomial is to use a prime number as your constant. Then make sure the
coefficient of x is not the sum of the factors of your prime number.
1. 3x3 +18x2 – 81x
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
3x is the GCF for all terms.
3x(x2 + 6x - 27)
Step 2: Factor the expression inside of
parenthesis.
We need a positive and negative number
whose product is -27 and whose sum is
6. (9 & -3) (9 must be positive since the
sum is positive)
Final Answer: 3x(x+9) (x - 3)
2. 2x3 – 8x2 – 8x
Step 1: Since the lead coefficient is greater than
1, see if there is a GCF for all terms.
2x is the GCF for all terms.
2x(x2- 4x - 4)
Step 2: Factor the expression inside of
parenthesis.
We need a positive and negative number
whose product is -4 and whose sum is -4.
The inside of the parenthesis cannot be
factored any further because there’s no
positive and negative number whose
product is -4 and whose sum is -4.
Final Answer: 2x(x2 – 4x – 4)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c
If you are factoring a trinomial with a lead coefficient that is greater than 1 and there is NO GCF to
factor out, then we will need to use a guess and check method in order to factor the trinomial.
Example 1
Factor: 2x2 +9x + 4
Example 2
Factor: 3x2 + 7x - 6
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Example 3
Factor: 4x2 -4x - 3
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c
Part 1: Factor completely.
1. 3x2 + 5x – 2
6. 4x2 + 4x +1
2. 2x2 – 3x – 2
7. 6x2 -5x - 6
3. 5x2 – 14x – 3
8. 4x2 – x - 3
4. 3x2 + 13x + 4
9. 36x2 + 12x + 4
5. 2x2 – 9x + 4
10. 8x2 – 14x + 6
Part 2: Mixed Factoring! Factor each trinomial completely using ANY method (or combination of
methods.)
1. x2 + 9x + 18
9. 24x2 + 80x + 24
2. 3x3 – 33x2 + 84x
10. 6x3 -24x2 -126x
3. 2x2 + 9x – 5
11. 5x3 -35x2 -40x
4. 10x2 + 5x – 15
12. 18x2 -21x -9
5. x2 –x – 56
13. 10x2 -32x +6
6. 6x2 – 10x – 4
14. 12x3 +36x2 + 27x
7. 3x2 – 33x + 54
15. 4x2 - 1
8. 2x3 + 20x2 + 42x
16. 27x2 -12
Factor Completely.
(2 points each)
1. 6x2 + 3x – 3
3. 24x2 +52x - 20
2. 8x2 +14x +6
4. 18x2 -12x + 2
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c - Answers
Part 1: Factor completely.
For all of the problems in Part 1, the lead coefficient is greater than 1. However, there is no GCF, so we must
use the guess and check method.
1. 3x2 + 5x – 2
6. 4x2 + 4x +1
Using the model (ax + b)(cx+d), we need:
Using the model (ax + b)(cx+d), we need:
ac=3 (3 &1 )and bd=-2 (-2 &1 or -1 &2)
ac=4 (4 &1 or 2 &2 )and bd=1 (1 &1)
(3x-2)(x+1) =3x2 +3x -2x -2 or
(4x+1)(x+1) = 4x2 +4x+x+1 or 4x2 +5x +1
3x2 +x - 2
(3x+1)(x-2) = 3x2 – 6x +x -2 or 3x2 -5x - 2
(2x+1)(2x+1) = 4x2 +2x +2x +1 or 4x2 +4x +1
(3x -1)(x+2) = 3x2 +6x – x -2 or 3x2 +5x – 2
(3x+2)(x-1) = 3x2 – 3x + 2x – 2 or 3x2 -x – 2
Final Answer: (2x+1)(2x+1) or (2x+1)2
Final Answer: (3x-1)(x+2)
2. 2x2 – 3x – 2
7. 6x2 -5x - 6
Using the model (ax + b)(cx+d), we need:
ac=2 (2 &1 )and bd=-2 (-2 &1 or -1 &2)
(2x-2)(x+1) = 2x2 +2x -2x -2 or 2x2 -2
(2x+1)(x-2) = 2x2 -4x +x -2 or
2x2 -3x -2
(2x -1)(x+2) = 2x2 +4x – x -2 or 2x2 +3x - 2
(2x+2)(x-1) = 2x2 -2x +2x - 2 or 2x2 -2
Final Answer: (2x+1)(x-2)
Using the model (ax + b)(cx+d), we need:
ac=6 (6 &1 or 3 & 2 )and bd=-6 (-6 &1 or -1 &6
or -3 &2 or -2 &3)
(6x-6)(x+1) = 6x2 +6x -6x -6 or 6x2 -6
(6x-1)(x+6) = 6x2 +36x –x -6 or 6x2 +35x - 6
(6x -3)(x+2) = 6x2 +12x -3x -6 or 6x2 +9x -6
(6x-2)(x+3) = 6x2 +18x -2x -6 or 6x2 +16x -6
(3x-6)(2x+1) = 6x2 +3x-12x -6 or 6x2 -9x -6
(3x-1)(2x+6) = 6x2 +18x -2x -6 or 6x2 +16x - 6
(3x -3)(2x+2) = 6x2 +6x -6x -6 or 6x2 -6
(3x-2)(2x+3) = 6x2 +9x -4x -6 or 6x2 +5x – 6
(3x+2)(2x-3) = 6x2 -9x +4x -6 or 6x2 -5x -5
Final Answer: (3x+2)(2x-3)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
3. 5x2 – 14x – 3
8. 4x2 – x – 3
Using the model (ax + b)(cx+d), we need:
Using the model (ax + b)(cx+d), we need:
ac=5 (5 &1 )and bd=-3 (-3 &1 or -1 &3)
ac=4 (4 &1 or 2 & 2)and bd=-3 (-3 &1 or -1 &3)
(5x-3)(x+1) = 5x2 +5x – 3x -3 or 5x2 +2x -3
(4x-3)(x+1) = 4x2 +4x – 3x -3 or 4x2 +x - 3
(5x+1)(x-3) = 5x2 -15x +x -3 or 5x2 -14x -3
After multiplying this problem, I notice that I need to
make x a negative x and then I would have
the correct factors. Therefore, if I try
(5x -1)(x+3) = 5x2 +15x –x -3or 5x2 +14x - 3
(5x+3)(x-1) = 5x2 -5x +3x -3 or 5x2 -2x -3
(4x +3) (x-1) I may be able to arrive at my
answer and not have to try all of the other
factors.
Final Answer: (5x+1)(x-3)
(4x+3)(x-1) = 4x2 -4x +3x -3 or 4x2 –x -3
Final Answer: (4x+3)(x-1)
4. 3x2 + 13x + 4
9. 36x2 + 12x + 4
Using the model (ax + b)(cx+d), we need:
ac=3 (3 &1 )and bd= 4 (2 &2 or 4 &1 or 1 & 4)
2
(3x+2)(x+2) =3x +6x +2x+4 or
2
3x +8x +4
The first thing I notice is that there is a GCF that I
can factor out. 4 is the GCF. We then have:
4(9x2 + 3x +1) Now I can factor inside of the
parenthesis:
(3x+4)(x+1) = 3x2 +3x +4x +4 or 3x2 7x +4
4(9x2 + 3x +1)
(3x +1)(x+4) = 3x2 +12x +x+4 or 3x2 +13x +4
Using the model (ax + b)(cx+d), we need:
ac=9 (9 &1 or 3 & 3 ) and bd= 1 (1 &1)
Final Answer: (3x+1)(x+4)
Since my sum is 3x, I am not going to start with 9
&1, I will start with 3 & 3.
(3x +1)(3x+1) = 9x2 + 3x +3x +1 or 9x2 + 6x +1
This cannot be factored, so our final answer is:
Final Answer: 4(9x2 + 3x +1)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
5. 2x2 – 9x + 4
Using the model (ax + b)(cx+d), we need:
ac=2 (2 &1 )and bd= 4 – Since we have -9x, the
factors of 4 must both be negative. (-2 &-2)(-4&-1)
(-1&-4)
(2x – 2)(x-2) = 2x2 -4x -2x +4 or 2x2 -6x +4
(2x -1)(x-4) = 2x2 -8x –x + 4 or 2x2 -9x +4
I can stop since I have found the factors.
Final Answer: (2x-1)(x-4)
10. 8x2 – 14x + 6
Notice that 2 is the GCF for all 3 terms. We’ll factor
a 2 out first.
2(4x2 – 7x + 3)
Using the model (ax + b)(cx+d), we need:
ac=4 (4 &1 )(2 & 2)and bd= 3. Since we have -7x,
the factors of 3 must both be negative
(-3 &-1)
(2x -3)(2x-1) = 4x2 -2x – 6x + 3 or 4x2 -8x +3
(4x-3)(x-1) = 4x2 – 4x – 3x + 3 or 4x2 – 7x + 3.
Final Answer: 2(4x-3)(x-1)
Part 2: Mixed Factoring! Factor each trinomial completely using ANY method (or combination of
methods.)
1. x2 + 9x + 18
9. 24x2 + 80x + 24
Since the lead coefficient is 1, I need to think of
two positive numbers that have a product of 18
and a sum of 9. (6&3)
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 8 is
the GCF, so first we’ll factor out an 8.
Final Answer: (x+6)(x+3)
8(3x2 + 10x + 3)
The lead coefficient inside of the parenthesis is
still greater than 1, so:
Using the model (ax + b)(cx+d), we need:
ac=3 (3 &1 )
and bd= 3.(3&1)(1&3)
Factor: 3x2 + 10x + 3
(3x +3) (x+1) = 3x2 + 3x+3x + 3 or 3x2 + 6x + 3
(3x+1)(x+3) = 3x2 + 9x + x + 3 or 3x2 + 10x + 3
Final Answer: 8(3x+1)(x+3)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
2. 3x3 – 33x2 + 84x
10. 6x3 -24x2 -126x
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 3x is
the GCF, so first we’ll factor out a 3x.
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 6x is
the GCF, so first we’ll factor out a 6x.
3x(x2 - 11x + 28)
6x(x2 - 4x - 21)
The lead coefficient inside of the parenthesis is 1,
so: We need two negative numbers whose
product is 28 and sum is -11. (-7 & -4)
Factor: x2 - 11x + 28
The lead coefficient inside of the parenthesis is 1,
so: We need a positive and negative number
whose product is -21 and whose sum is -4. (-7&3
- Since 4 is negative, the largest number (7) must
be negative.
(x-7)(x-4)
Factor: x2 - 4x -21
Final Answer: 3x(x-7)(x-4)
(x -7)(x+3)
Final Answer: 6x(x-7)(x+3)
3. 2x2 + 9x – 5
11. 5x3 -35x2 -40x
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. There
is no GCF.
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. The
GCF is 5x, so I will first factor out 5x.
Using the model (ax + b)(cx+d), we need:
5x(x2 – 7x – 8)
ac=2 (2 &1 )
Now the lead coefficient inside of the parenthesis
is 1, so I will factor: x2 – 7x – 8. I need a positive
and negative number whose product is -8 and
whose sum is -7. (-8 & 1) (8 must be negative
since 7 is negative.
and bd= -5.(-5&1)(-1&5)
(2x – 5)(x+1) = 2x2 +2x – 5x -5 or
2x2 -3x – 5
(2x – 1)(x +5) = 2x2 + 10x – x -5 or 2x2 +9x -5
(x – 8)(x+1)
Final Answer: (2x-1)(x+5)
Final Answer: 5x (x-8)(x+1)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
4. 10x2 + 5x – 15
12. 18x2 -21x -9
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. The
GCF is 5.
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 3 is
the GCF. We’ll first factor out a 3.
5(2x2 + x – 3)
3(6x2 – 7x – 3)
The lead coefficient inside of the parenthesis is
greater than 1, so:
Since the lead coefficient inside of the parenthesis
is greater than 1, we need to:
Using the model (ax + b)(cx+d), we need:
Use the model (ax + b)(cx+d), we need:
ac=2 (2 &1 )
ac=6 (2 &3)(6&1)
and bd= -3.(-3&1)(-1&3)
(2x –3)(x+1) = 2x2 +2x – 3x -3 or
2x2 -x – 3
We need a sum of positive x, so let’s switch the
signs:
2
(2x+3)(x-1) = 2x – 2x + 3x – 3 or
Factor: 6x2 – 7x – 3
(3x -3)(2x +1) = 6x2 +3x – 6x – 3 or 6x2 – 3x – 3
(3x+1)(2x-3) = 6x2 – 9x + 2x -3 or 6x2 -7x -3
2
2x +x - 3
Final Answer: 5(2x+3)(x-1)
Final Answer: 3(3x+1)(2x-3)
5. x2 –x – 56
Since the lead coefficient is 1, we need to find a
positive and negative number whose product is 56 and whose sum is -1. (-8&7) (Since x is
negative, our largest number (8) must be negative
because -8 + 7 = -1.
(x-8)(x+7)
and bd= -3.(-3&1)(-1&3)
13. 10x2 -32x +6
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 2 is
the GCF.
2(5x2 – 16x + 3)
The lead coefficient inside of the parenthesis is
still greater than one, so:
Using the model (ax + b)(cx+d), we need:
Final Answer: (x-8)(x+7)
ac=5 (5 &1 ) and bd= 3.(-3&-1)(-1&-3)(Both
factors of 3 must be negative since 16 is negative)
(5x – 1)(x-3) = 5x2 – 15x – x +3 or 5x2 -16x +3
Final Answer: 2(5x-1)(x-3)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
6. 6x2 – 10x – 4
14. 12x3 +36x2 + 27x
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 2 is
the GCF.
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 3x is
the GCF.
2(3x2 – 5x – 2)
3x(4x2 + 12x + 9)
The lead coefficient inside of the parenthesis is
greater than 1, so:
The lead coefficient inside of the parenthesis is
greater than 1, so:
Using the model (ax + b)(cx+d), we need:
Using the model (ax + b)(cx+d), we need:
ac=3 (3 &1 )
ac=4 (2 &2)(4&1)
and bd= -2.(-2&1)(-1&2)
(3x + 1)(x – 2) = 3x2 – 6x + x – 2 or 3x2 – 5x – 2
and bd= 9.(9&1)(3&3)
I know that 2 x 3 = 6 and 6+6 = 12, so let’s try 2
&2 for ac, and 3 & 3 for bd.
(2x+3)(2x+3) = 4x2 + 6x +6x +9 or 4x2 + 12x + 9
Final Answer: 2(3x+1)(x-2)
Final Answer: 3x(2x+3)(2x+3) or 3x(2x+3)2
7. 3x2 – 33x + 54
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 3 is
the GCF.
3(x2 – 11x + 18)
Now we will factor inside of the parenthesis. I
need two negative numbers whose product is 18
and sum is -11. (-9&-2)
15. 4x2 – 1
Since the middle term is missing, I know that this
is a difference of two squares. What number can
you square to get 4 (2) and what number can you
square to get 1 (1)?
(2x+1)(2x-1)
Multiply to check: 4x2 – 2x + 2x – 1 or 4x2 -1
(x -9)(x-2)
Final Answer: (2x +1)(2x-1)
Final Answer: 3 (x-9)(x-2)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
8. 2x3 + 20x2 + 42x
16. 27x2 -12
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 2x is
the GCF.
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for both terms. 3 is the
GCF.
2x(x2 + 10x + 21)
3(9x2 – 4)
Now factor inside of the parenthesis. The lead
coefficient is 1, so we need two positive numbers
whose product is 21 and whose sum is 10. (7&3)
Inside of the parenthesis we have no middle term,
so we have a difference of two squares. What
can we square to get 9 (3) and what can we
square to get 4 (2)?
(x+7)(x+3)
(3x +2)(3x-2)
Final Answer: 2x (x+7)(x+3)
Final Answer: 3(3x+2)(3x-2)
Factor Completely.
(2 points each)
1. 6x2 + 3x – 3
3. 24x2 +52x – 20
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 3 is
the GCF.
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 4 is
the GCF.
3(2x2 + x - 1)
4(6x2 + 13x – 5)
Now factor inside of the parenthesis. The lead
coefficient is greater than 1, so:
(Now factor inside of the parenthesis. The lead
coefficient is greater than 1, so:
Using the model (ax + b)(cx+d), we need:
Using the model (ax + b)(cx+d), we need:
ac=2 (2 &1)
ac=6 (6 &1)(3&2)
and bd= -1.(1&-1)(-1&1)
and bd= -5.(5&-1)(-1&5)
Factor: 2x2 + x – 1
Factor: 6x2 + 13x - 5
(2x -1)(x+1)= 2x2 +2x – x -1 or 2x2 + x -1
(3x -1)(2x + 5) = 6x2 + 15x -2x – 5 or 6x2 + 13x -5
Final Answer: 3 (2x-1)(x+1)
Final Answer: 4(3x-1)(2x+5)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
2. 8x2 +14x +6
4. 18x2 -12x + 2
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 2 is
the GCF.
Since the lead coefficient is greater than 1, I will
first see if there is a GCF for all three terms. 2 is
the GCF.
2(4x2 +7 x +3)
2(9x2 -6x +1)
Now factor inside of the parenthesis. The lead
coefficient is greater than 1, so:
Now factor inside of the parenthesis. The lead
coefficient is greater than 1, so:
Using the model (ax + b)(cx+d), we need:
Using the model (ax + b)(cx+d), we need:
ac=4 (4 &1)(2&2)
ac=9 (9 &1)(3&3)
and bd= 3.(1&3)(3&1)
and bd= 1.(-1&-1)
Factor: 4x2 +7 x+ 3
Factor: 9x2 – 6x +1
(4x +3)(x+1) = 4x2 +4x +3x+3 or 4x2 +7x +3
(3x -1)(3x-1) = 9x2 -3x -3x +1 or 9x2 – 6x +1
Final Answer: 2(4x+3)(x+1)
Final Answer: 2 (3x-1)(3x-1) or 2(3x-1)2
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Unit 9: Factoring Chapter Test
Part 1: Factor Completely. Remember to factor out the greatest common factor first if it is other
than 1.
1. x2 + 12x + 32
7. 3y2 + 21y + 36
2. y2 – 2y – 35
8. 2x3 – 18x
3. n2-8n + 16
9. 12x4y2 + 18x3y3 – 24x2y4
4. x2 – 13x + 36
10. 6x2 + 5x - 4
5. 8x4 + 2x2 – 4x
11. 4p2 – 7p - 2
6. 2x3 + 6x2 – 20x
12. 4x2 + 10x – 6
Part 2: Answer each problem completely.
1. What polynomial when factored gives: (3x+ 2y)( 2x – 5y)
2. One factor of the polynomial x2 + 27x + 180 is (x+15). What is the other factor?
3. Find two values for b. Explain how you determined your answer.
x2 + bx – 12
4. Find a value for c that will make this a perfect square trinomial.
x2 + 18x + c
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Unit 9: Factoring Chapter Test – Answer Key
Part 1: Factor Completely. Remember to factor out the greatest common factor first if it is other
than 1. (1 point each)
1. x2 + 12x + 32
7. 3y2 + 21y + 36
We need to find two positive integers whose
product is 32 and whose sum is 12.
8&4
8●4 = 32
8+4 = 12
Since the lead coefficient is greater than1, we
need to check to see if there is a GCF that can
be factored out. The GCF is 3.
3(y2 + 7y + 12)
Final Answer: (x+8)(x+4)
Now we need to factor the polynomial inside of
the parenthesis. We need two positive
numbers whose product is 12 and whose sum
is 7.
4&3
Final Answer: 3(x+4)(x+3)
2. y2 – 2y – 35
8. 2x3 – 18x
We need to find two integers whose product is
-35 and whose sum is -2. Since the product is
negative, one integer is positive and one
negative. The larger integer is negative since
the sum is negative.
Since the lead coefficient is greater than1, we
need to check to see if there is a GCF that can
be factored out. The GCF is 2x.
-7 & 5
Now we need to factor the polynomial inside of
the parenthesis. Since there is no middle term
and 9 is a perfect square, we know this factors
as a difference of two squares:
-7●5 = -35
-7+5 = -2
Final Answer: (y-7)(y+5)
2x(x2 - 9)
Final Answer: 2x(x+3)(x-3)
3. n2-8n + 16
9. 12x4y2 + 18x3y3 – 24x2y4
We need to find two integers whose product is
16 and whose sum is -8. Since the product is
positive and the sum is negative, both integers
must be negative.
First we need to see if there is a GCF. The
GCF is 6x2y2. Factor out 6x2y2.
-4 & -4
There is no way to factor inside of the
parenthesis. Therefore:
-4●(-4) = 16
-4+(-4) = -8
Final Answer: (n-4)(n-4) or (n-4)2
6x2y2 (2x2+ 3xy – 4y2)
Final Answer: 6x2y2 (2x2+ 3xy – 4y2)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
4. x2 – 13x + 36
10. 6x2 + 5x – 4
We need to find two integers whose product is
36 and whose sum is -13. Since the product is
positive and the sum is negative, both integers
must be negative.
Since the lead coefficient is greater than 1, look to
see if there is a GCF. There is no GCF for this
polynomial, so we must use the guess and check
method for factoring.
-9 & -4
Factors of 6: 6,1 & 3,2
-9●(-4) = 36
-9+(-4) = -13
Final Answer: (x-9)(x-4)
Factors of -4: (2,-2) & (-4,1) & (-1,4)
(2x -1) (3x+4) = 6x2 + 8x – 3x – 4 or 6x2 +5x -4
Final Answer: (2x -1) (3x+4)
5. 8x4 + 2x2 – 4x
First we need to check to see if there is a GCF
that can be factored out. 2x is the GCF.
2x(4x3 + x – 2)
The inside of the parenthesis cannot be
factored any further, therefore:
Final Answer: 2x(4x3 + x – 2)
11. 4p2 – 7p – 2
Since the lead coefficient is greater than 1, look to
see if there is a GCF. There is no GCF for this
polynomial, so we must use the guess and check
method for factoring.
Factors of 4: 2,2 & 4,1
Factors of -2: (1,-2) & (-2,1)
(4p +1) (p-2) = 4p2 – 8p + p – 2 or 4p2 -7x -2
Final Answer: (4p +1) (p-2)
6. 2x3 + 6x2 – 20x
12. 4x2 + 10x – 6
First we need to check to see if there is a GCF
that can be factored out. 2x is the GCF.
Since the lead coefficient is greater than 1, look to
see if there is a GCF. 2 is the GCF.
2x(x2 + 3x - 10)
2(2x2 + 5x – 3)
Now we must factor the inside of the
parenthesis. We need a positive and negative
integer whose product is -10 and whose sum is
3. The larger integer will be positive since the
sum is positive. 5 & -2
Now factor the inside of the parenthesis using the
guess and check method.
Final Answer: 2x(x+5)(x-2)
2(2x-1) (x+3) =2(2x2 +6x - x – 3) or 2(2x2 +5x -3)
Factors of 2: 2,1
Factors of -3: (1,-3) & (-3,1)
Final Answer: 2(2x-1)(x+3)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Part 2: Answer each problem completely. (2 points each)
1. What polynomial when factored gives: (3x+ 2y)( 2x – 5y)
We can multiply to figure out the polynomial: Use the foil method.
(3x+2y)(2x-5y)
3x(2x) + 3x(-5y) + 2y(2x) + 2y(-5y)
6x2 – 15xy + 4xy – 10y2
Use FOIL
Multiply the terms
6x2 – 11xy – 10y2
Simplify the middle terms.
Final Answer: 6x2 – 11xy – 10y2
2. One factor of the polynomial x2 + 27x + 180 is (x+15). What is the other factor?
We know that our two chosen integers must have a product of 180 and a sum of 27. Since we know
that one factor is (x+15), we know that one of our integers is 15. The other integer must be 12 since
180/15 = 12. Also, 15+ 12 = 27, so this confirms our choice.
Therefore the other factor is: (x+ 12)
Final Answer: (x+12)
3. Find two values for b. Explain how you determined your answer.
x2 + bx – 12
We know that we need two integers, one positive and one negative, whose product is -12. The sum
of these two integers is the variable b in the polynomial.
Integers whose product is -12: (-4)(3); (-3)(4); (-6)(2); (-2)(6); (-12)(1); (12)(-1)
b could be equal to: -1, 1, -4, 4,-11, 11 (If you add the integers together, you will end up with their
sum, which is the variable b.)
4. Find a value for c that will make this a perfect square trinomial.
x2 + 18x + c
Letter C must be a perfect square. The square root of this number multiplied by 2 is 18. The only
number that can represent letter c is 81. The factors would be (x+9)2
This test is worth 20 points
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Cumulative Test on Polynomials and Factoring
Part 1: Complete each problem by circling the correct answer.
For problems 1-8, simplify the expression. State the answer in standard form.
1. (3x2+8x – 11) + (2x2 – 6x +4)
A. 5x2 + 14x + 15
C. 5x2 + 2x – 7
B. 5x2 + 2x + 7
D. 5x2 + 2x – 15
2. (4x4 + 2x2 – 3x + 1) + (3x3 – 2x2 + 8x – 6)
A. 4x4 + 3x3 + 5x – 5
C. 4x4 + 3x2 + 4x2 – 3x +8x – 5
B. 7x7 + 5x – 5
D. 4x4 + 3x3 + 2x2 + 5x – 5
3. (6x2 – 3x + 9) – (4x2 + 8x – 6)
A. 2x2 + 5x + 3
C. 2x2 + 11x + 15
B. 2x2 – 11x + 15
D. 10x2 + 5x + 3
4. 3(2x2 – 4x +2) – 2(x2 + 4)
A. 4x2 – 12x + 14
C. 8x2 – 12x – 2
B. 8x2 – 12x + 14
D. 4x2 – 12x – 2
5. (5r-3)2
A. 25r2 – 9
C. 5r2 + 9
B. 25r2 – 30r + 9
D. 25r2 + 9
6. (x+6)(x-6)
A. x2 + 12x – 36
C. x2 + 36
B. x2 – 36
D. x2 -6x – 36
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
7. (3x-2)(2x+1)
A. 6x2 + 7x – 2
C. 6x2 – x + 2
B. 6x2 + x – 2
D. 6x2 – x – 2
8. (x+4)2(3x-2)
A. 3x3 + 22x2 +32x – 32
C. 3x2 + 10x – 8
B. 3x3 – 2x2 + 48x – 32
D. 3x3 + 26x2 + 48x – 32
For problems 9-10, factor the polynomial.
9. x2 -3x -70
A. (x+10)(x-7)
C. (x+7)(x-10)
B. (x+5)(x-14)
D. (x+14)(x-5)
10. 6x2 – 2x – 8
A. (3x-4)(2x+2)
B. (3x+2)(2x-4)
C. (2x-2)(3x+4)
D. (2x+4)(3x-2)
Part 2: Short Answer
1. A square table measures 3x +2y units. Write an expression in simplified form that represents the
area of the table.
2. A rectangular sheet of paper has a width of x units. The area of the paper is x2+8x. Write an
expression that represents the length of the rectangular sheet of paper.
3. Factor the polynomial completely.
2x2 + 11x + 12
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
4. Factor the polynomial completely.
9x3 +6x2 -48x
5. Factor the polynomial completely.
y2 – 121
6. A rectangle has the following dimensions: width: (2x + y)
length: (5x – 2y)
Express the area of the rectangle as a polynomial in standard form.
7. A triangle has the following dimensions: Side 1: x2 +2x -1
Side 2: x2 + 2x -1
Side 3: 2x2 +x – 3
Write an expression that represents the perimeter of the triangle.
8. A rectangle has a length of 8x -4 and a width of 4x +3. The perimeter of the rectangle is 58 units.
Find the width of the rectangle.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Part 3: Extended Response.
1. You have a rectangular piece of wood in which
you are making a carnival game. You plan
to cut out 2 circles that children can use to try to
throw a bean bag through at the carnival.
Complete each bullet below given
the following dimensions:
Length of rectangular piece of wood: (5x+2)
Width of rectangular piece of wood: (3x+1)
Radius of each circle: (x+1)
•
Write an algebraic expression for the area of the
rectangular piece of wood before the circles are cut.
•
•
Write an algebraic expression for the area of 1 circle.
Write an algebraic expression for the area of the wood after both circles are cut out. Explain
how you determined your answer.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Cumulative Test on Polynomials and Factoring - Answer Key
Part 1: Complete each problem by circling the correct answer. (1 point each)
For problems 1-8, simplify the expression. State the answer in standard form.
1. (3x2+8x – 11) + (2x2 – 6x +4)
Write the expression vertically.
3x2 + 8x – 11
+ 2x2 – 6x + 4
5x2 +2x - 7
A. 5x2 + 14x + 15
C. 5x2 + 2x – 7
B. 5x2 + 2x + 7
D. 5x2 + 2x – 15
2. (4x4 + 2x2 – 3x + 1) + (3x3 – 2x2 + 8x – 6)
Write the expression vertically.
4x4 + 0x3 +2x2 – 3x + 1
+
3x3 -2x2 + 8x – 6
4x4 + 3x3 +0x2 +5x - 5
A. 4x4 + 3x3 + 5x – 5
C. 4x4 + 3x2 + 4x2 – 3x +8x – 5
B. 7x7 + 5x – 5
D. 4x4 + 3x3 + 2x2 + 5x – 5
3. (6x2 – 3x + 9) – (4x2 + 8x – 6)
Step 1: Rewrite as an addition problem.
(6x2 – 3x + 9) + (-4x2 – 8x + 6)
Write the new expression vertically.
6x2 – 3x + 9
+ -4x2 – 8x + 6
2x2 – 11x + 15
A. 2x2 + 5x + 3
C. 2x2 + 11x + 15
B. 2x2 – 11x + 15
D. 10x2 + 5x + 3
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
4. 3(2x2 – 4x +2) – 2(x2 + 4)
Step 1: Distribute the 3 throughout the first set of parenthesis and the -2 throughout the second set of
parenthesis:
(6x2 – 12x + 6) + (-2x2 – 8)
Step 2: Write the new expression vertically.
6x2 – 12x + 6
+ -2x2 +0x -8
4x2 – 12x -2
A. 4x2 – 12x + 14
C. 8x2 – 12x – 2
B. 8x2 – 12x + 14
D. 4x2 – 12x – 2
5. (5r-3)2
Step 1: Use the Difference of a Square rule.
(a-b)2 = a2 – 2ab + b2
(5r-3)2 = (5r)2 – 2(5r)(3) + 32
(5r-3)2 = 25r2 – 30r +9
A. 25r2 – 9
C. 5r2 + 9
B. 25r2 – 30r + 9
D. 25r2 + 9
6. (x+6)(x-6)
Step 1: Use the Difference of Two Square Rule:
(a+b)(a-b) = a2 – b2
(x+6)(x-6) = x2 – 62
(x+6)(x-6) = x2 - 36
A. x2 + 12x – 36
C. x2 + 36
B. x2 – 36
D. x2 -6x – 36
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
7. (3x-2)(2x+1)
Step 1: Using FOIL multiply the two binomials.
(3x-2)(2x+1) = 3x(2x) + 3x(1) + (-2)(2x)+ (-2)(1)
6x2 + 3x – 4x – 2
Step 2: Combine like terms. 6x2 – x – 2
Final Answer: 6x2 – x - 2
A. 6x2 + 7x – 2
C. 6x2 – x + 2
B. 6x2 + x – 2
D. 6x2 – x – 2
8. (x+4)2(3x-2)
Step 1: Using the Square of A sum rule, multiply (x+4)2
(x+4)2 = x2 + 8x + 16
(a2 + 2ab + b2)
Step 2: Multiply (3x-2) (x2 + 8x + 16)
Use the extended distributive property.
3x(x2) + 3x(8x) + 3x(16) + (-2)(x2) + (-2)(8x) + (-2)(16)
3x3 + 24x2 + 48x – 2x2 - 16x – 32
Step 3: Rewrite with like terms together.
3x3 + 24x2 – 2x2 + 48x – 16x – 32
Step 4: Combine like terms:
3x3 + 22x2 + 32x - 32
A. 3x3 + 22x2 +32x – 32
C. 3x2 + 10x – 8
B. 3x3 – 2x2 + 48x – 32
D. 3x3 + 26x2 + 48x – 32
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
For problems 9-10, factor the polynomial.
9. x2 -3x -70
Since the lead coefficient is 1, we need to think of two numbers that have a product of -70 and a sum
of -3. One integer must be positive and one must be negative. The larger integer will be negative
since the sum is -3.
**Since this is multiple choice – we should take a look at the answers given. The answers are using
factors of 10 and 7 and 5 and 14. Of those two choices, let’s see which one works.
***I know that the larger integer must be negative, since the sum is -3; therefore, I can eliminate A
because the larger integer, 10 is positive. For the same reason, I can eliminate D. Let’s try letter C.
We’ll multiply: (x+7)(x-10) = x2 – 10x + 7x – 70
x2 – 3x – 70
This is the answer we need; therefore, letter C is the correct choice.
A. (x+10)(x-7)
C. (x+7)(x-10)
B. (x+5)(x-14)
D. (x+14)(x-5)
10. 6x2 – 2x – 8
Since this is multiply choice, we should use the answers to our advantage. It may be best to guess
one of the answer choices below and check by multiplying. We would have to use
the guess and check strategy anyway to factor this polynomial since the lead
coefficient is greater than 1.
A: 2(3x-4)(x+1)
(6x – 8)(x+1)
- Distribute the 2 throughout the first parenthesis
6x2 +6x – 8x -8
= 6x2 – 2x – 8
Our first try works!!!
A. 2(3x-4)(x+1)
B. 2(x-2) (3x+2)
C. 2(x-1)(3x+4)
D. 2(x+2)(3x-2)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Part 2: Short Answer (2 points each)
1. A square table measures 3x +2y units. Write an expression in simplified form that represents the
area of the table top.
A square table has four sides with the same measurement (3x+2y).
Area of a square is: A = l● w. or A = s2
A = (3x+2y)2
We can use the Square of a Sum rule in order to multiply and simplify this expression.
(a+b)2 = a2 + 2ab +b2
(3x+2y)2 = (3x)2 + 2(3x)(2y) + (2y)2
A = 9x2 + 12xy + 4y2
2. A rectangular sheet of paper has a width of x units. The area of the paper is x2+8x. Write an
expression that represents the length of the rectangular sheet of paper.
width= x
length = ?
Area = x2 + 8x
The formula for area of a rectangle is: A = lw
Substitute our given information into this formula:
x2 + 8x = Lx
Now we need to solve for L by dividing by x.
x2 + 8x /x = l/x
****You may also, have solved
this problem mentally by
thinking:
2
x + 8x = L
x
An x can be factored out of the numerator: x(x+ 8) = L
X
Now we can simplify the fraction: x(x+ 8) = L
X
x + 8 = L We end up with an expression of x + 8 = L
x(?) = x2 + 8x
What can I
multiply by x in order to get
x2+ 8x?
x(x +8) = x2 + 8x
Length = x+8
Let’s check:
A = lw
A = (x+8)(x)
A = x2 + 8x
This expression matches the original problem; therefore, our answer is correct
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
3. Factor the polynomial completely.
2x2 + 11x + 12
Factors of 2: (2●1)
Factors of 12: (12 ●1) (6●2) (3●4)
Guess and check: We need the inside terms to add up to 11.
(2x + 6)(x+2) = 2x2 + 4x + 6x + 12
= 2x2+ 10x + 12
(2x + 3)(x+4) = 2x2 + 8x + 3x + 12 = 2x2 + 11x + 12
Final Answer: (2x+3)(x+4)
4. Factor the polynomial completely.
9x3 +6x2 -48x
First factor out the GCF which is 3x.
3x(3x2 + 2x – 16)
Now factor: 3x2 + 2x – 16
Factors of 3: 3,1
Factors of -16: -4(4), -8(2) or -2(8)
We must use the guess and check method:
3x (3x+8) (x-2)
Let’s check: (3x+8) (x-2) = 3x2 – 6x + 8x – 16 = 3x2 + 2x – 16 (Yeah ! This worked)
Final answer: 3x(3x+8)(x-2)
5. Factor the polynomial completely.
y2 – 121
Since y2 and 121 are both perfect squares and there is no middle term, I recognize this product as a
Difference of Two Squares. I know that 112 = 121. Therefore, my factors are:
(y+11)(y-11)
Final Answer: (y+11)(y-11)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
6. A rectangle has the following dimensions: width: (2x + y)
length: (5x – 2y)
Express the area of the rectangle as a polynomial in standard form.
The formula for area of a rectangle is: A = lw
A = (2x+y)(5x-2y)
Use FOIL to multiply.
A = 2x(5x) +2x(-2y) + y(5x) + y(-2y)
A = 10x2 – 4xy + 5xy – 2y2
A = 10x2 + xy – 2y2
Final Answer: 10x2 + xy – 2y2
7. A triangle has the following dimensions: Side 1: x2 +2x -1
Side 2: x2 + 2x -1
Side 3: 2x2 +x – 3
Write an expression that represents the perimeter of the triangle.
To find the perimeter of the triangle, we must add all sides together. I will set it up vertically.
x2 + 2x – 1
x2 + 2x – 1
+ 2x2 + x – 3
4x2 + 5x – 5
The perimeter of the triangle is: 4x2 + 5x - 5
8. A rectangle has a length of 8x -4 and a width of 4x +3. The perimeter of the rectangle is 58 units.
Find the width of the rectangle.
The formula for perimeter of a rectangle is: P = 2L + 2w
P = 2(8x-4) + 2(4x+3)
P = 16x – 8 + 8x + 6
Distribute.
P = 16x + 8x – 8 + 6
Write like terms together.
P = 24x – 2
Expression that represents the perimeter.
58 = 24x – 2
Now substitute 58 for P since we know the perimeter is 58 units.
58 +2 = 24x -2 + 2
Add 2 to both sides
60 = 24x
60/24 = 24x/24
Divide by 24 on both sides.
2.5 = x
This is the value for x
4(2.5) + 3 = 13
Substitute 2.5 for x into the expression for width (4x+3) The width is 13 units.
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
Part 3: Extended Response. (4 points)
1. You have a rectangular piece of wood in which
you are making a carnival game. You plan
to cut out 2 circles that children can use to try to
throw a bean bag through at the carnival.
Complete each bullet below given
the following dimensions:
Length of rectangular piece of wood: (5x+2)
Width of rectangular piece of wood: (3x+1)
Radius of each circle: (x+1)
•
Write an algebraic expression for the area of the
rectangular piece of wood before the circles are cut.
Area of a rectangle: A = lw
A = (5x+2)(3x+1)
A = 15x2 + 5x + 6x + 2
or A = 15x2 + 11x + 2
Use FOIL to multiply
The Area of the Wood before circles are cut is: 15x2 + 11x + 2
•
Write an algebraic expression for the area of 1 circle.
The formula for area of a circle is: A = πr2
A = π (x+1)2
A = π (x2 + 2x+ 1)
Use square of a sum rule to multiply the expression.
The area of one circle is: π (x2 + 2x+ 1) or 3.14(x2 + 2x + 1)
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com
Unit 9: Factoring
•
Write an algebraic expression for the area of the wood after both circles are cut out. Explain
how you determined your answer.
Step 1: We need to find the expression that represents the total area of both circles that will be cut
out. Since the area of 1 circle is: 3.14(x2 + 2x + 1) , we need to multiply this by 2.
3.14(2)(x2 + 2x +1) or 6.28(x2 + 2x + 1)
Step 2: Let’s distribute the 6.28
6.28x2 + 12.56x + 6.28
This represents the total area of both circles that will be cut out.
Step 3: Subtract the area of the circles from the total area of the wood.
15x2 + 11x + 2 – (6.28x2 + 12.56x + 6.28)
Step 4: Rewrite as an addition problem.
15x2 + 11x + 2 + (-6.28x2 - 12.56x - 6.28)
Step 4: Add. Set up vertically.
15x2 + 11x + 2
+ -6.28x2 – 12.56x – 6.28
8.72x2 – 1.56x - 4.28
The expression that represents the area of the board after two circles are cut is:
8.72x2 – 1.56x - 4.28. I first multiplied the expression that represents the area of the circle
by 2 since there are two circles. Then I subtracted the expression that represents the area of
two circles, from the area of the rectangle.
This test is worth 30 points
Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com