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Thus far, we have learned the following involving QUADRATICS:
We have studied LINEAR equations ---- now, it’s time to move to another family . . .
The term QUADRATIC is from the Latin for “square” – due to having x2.
Standard Form:
Ax2 + Bx + C = 0 / Ax2 + Bx + C = y / Ax2 + Bx + C = f(x)
(all OK)
The graph of this form of equation results in a conic section known as a PARABOLA.
If A is positive, the parabola OPENS UP . . . the VERTEX is a MINIMUM.
If A is negative, the parabola OPENS DOWN . . . the VERTEX is a MAXIMUM.
If A is zero ---- then it’s not quadratic.
The AXIS OF SYMMETRY is a vertical line that splits the PARABOLA into 2 equal parts.
The EQUATION of the AXIS OF SYMMETRY is x = - b / 2a.
The VERTEX must lie on the AOS. The x-coordinate of the vertex is x = - b / 2a.
To find the y-coordinate of the VERTEX, plug in the x-coordinate into the original equation.
The y-intercept is the coordinates (0, C). The x-intercept(s) can be a bit more complicated . . .
To graph a quadratic
1. Get in standard form.
2. Graph the AOS.
3. Graph the vertex.
4. Graph the y-intercept.
5. “Mirror” the point across from the y-intercept.
Finally, when asked to SOLVE a QUADRATIC, you must find the:
x-intercept(s), root(s), solution(s) . . . these are all SYNONYMOUS
Sometimes, you can find these SOLUTIONS graphically, or using the TABLE feature on your
calculator; however, these SOLUTIONS can also be found by FACTORING the QUADRATIC.
There are several methods in which this can be accomplished.
This packet introduces you to the process of FACTORING QUADRATICS.
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ALGEBRA II – CHAPTER FIVE – FACTORING QUADRATICS
Definitions:
Prime number:
A whole number that has factor only of itself and 1.
Composite number: A whole number greater than 1, that has more than 2 factors.
Prime factorization: The expression of a whole number as the product of prime factors.
i.e. 90 = 2 • 45 = 2 • 3 • 15 = 2 • 3 • 3 • 5
You can get this by the ladder method you learned in pre-algebra. Start dividing by prime numbers until it cannot
de divided any more.
90
45
15
5
1
And the factors are 2•3•3•5
2
3
3
5
1. Find the prime factorization of:
a. 39
b. -100
c. 18 x 2
☺1
d. 35 c 2 d 3
e. 550 e3 d 4
f. 160 pg2
Find the GCF (Greatest common factor)
The Greatest common factor of 2 or more monomials is the is the product of their common factors.
(If the GCF is 1, then the two numbers are relatively prime)
You can use the ladder method to find the GCF: ex: Find the GCF of 48 and 60.
The GCF of the two numbers is 2•2•3
2
2
3
48
24
12
4
60
30
15
5
2. Find the GCF of:
a. 10 and 15
e. 18 xy and 36 y2
☺2
b. 54 and 63
c. 27 and 72
f. 12 qr and 8r2 and 16 rs
g. 15 r2s and 35 s2 and 70 rs
h. 28 q2b2 and 63b3 and 91 b3
d. 15 and 45 and 60
i. 14 m2n2 and 18 mn and 2 m2n2
3. Twin primes are primes that are 2 consecutive odd numbers that are prime. The first example of a set of
twin primes is 3 and 5. Can you find the next 5 pairs?
2
Factoring Using the Distributive Property
Example: Use the Distributive property to factor the following: 12 a2 + 16 a
Find the GCF of 12 a2 and 16 a. Then write the answer as the product of the GCF and the remaining factors.
i.e. The GCF is 4a. Therefore:
12 a2 + 16 a = 4a ( 3a ) + 4a ( 4 ) = 4a ( 3a + 4) Using the distributive property
☺3
Now try these:
1. 18 cd2 + 12 c2d + 9 cd
3. a5b - a
2. 5x + 30 y
This is the first, and most basic, form of factoring.
It is called Removing the Common factor. You always use this method FIRST if there is a common factor.
Method One:
CM
Common monomial factorization: This method should always be tried first to make other methods of
factorization easier. This method could be called distributive factorization because it reverses the
distributive property.
Explanation: Using the distributive property you would multiply 6( 2x –3) by multiplying the factors 6
and 2x together (12x) and add to your answer what you would get by multiplying the factors 6 and –3
together (-18). Therefore 6(2x – 3) = 12x – 18. So if you were asked to factor 12x – 18, the method used
would be Common Monomial Factorization because this method reverses the distributive process. THE
PROCESS IN WORDS: Find the greatest common factor (GCF) in all terms, divide all terms by that factor
and write the divisor and the dividend as factors. EX: Since 6 is the GCF of 12x and –18, I will divide
both terms by 6,the divisor, and get (2x – 3), the dividend. Therefore the factors of 12x –18 are 6(2x – 3).
Ex 1: Factor: 12 + 30 x = 6 (2 + 5x)
Ex 2: Factor: 8 n – 4n2 + 12n4 = 4n(1- n + 3n3)
☺4
Exercises: Using CM, factor the following:
Take out the Greatest Common Factor. Check by using the distributive property to multiply the factors together.
This factorization is called Common Monomial Factoring (CM). If nothing can be removed, it is PRIME.
1. 11x + 44x2y
2. 14xz - 18xz2
3. 18xy2 - 24x2y
4. 25a2b2 + 30ab3
5. 2m3n2 - 16m2n 3+ 8mn
6. 3x3y + 9xy2 + 36xy
7. 28a2b2c2 + 21a2bc2 - 14abc
8. 12ax + bx + 32cx
9. a + a2b + a3b3
10. 27ab2 + 9b
11. c4 + c6 - c2
12. 6a2 - 6ab - 3ca
13. 3my -ab + by
14. 24a2 - 9ab + 32ab
15. y - 9aby + 3by
16. 12a2 - 9ab + 24a
Method 2 - Factoring By Grouping
FG
Factoring by Grouping: This is recognized because it usually (but not always) involves four terms. Note
that all expressions that contain four terms cannot be factored in this manner. To accomplish this
3
factorization, look for ways in which the terms can be ordered and grouped so that they contain common
binomial factors. It is an extension of the CM factoring.
Ex. 1: x2 - 8x + 7x – 56
= (x2 - 8x) + (7x – 56 )
= x(x – 8) + 7(x – 8)
= (x – 8) (x + 7)
Group
CM Factoring of each group
Pulling out the common binomial factor
Ex 2. ax + ay + bx + by 56
= (ax + ay) + (bx + by )
= a(x + y) + b(x + y)
= (x + y) ( a + b)
Group
CM Factoring of each group
Pulling out the common binomial factor
Ex 3. 4hk2 – 4k3 – 9h3 + 9h2k = (4hk2 – 4k3) – (9h3 - 9h2k)
= 4k2(h – k) – 9h2(h – k)
= (h – k)(4k2 – 9h2)
Grouping (Note the change of sign
because a negative was factored out.)
CM Factoring of each group
Pulling out the common binomial factor
In the next example, reordering must occur before grouping for factorization to be successful. This is complicated
but, fortunately, rare. Perhaps the best clue is to look at the coefficients first, to see if they have anything in
common.
Ex. 4: 6a2 + 4b – 3a – 8ab
=
=
=
=
Ex 5: 5a2 - 4ab + 12b3 - 15ab2
=
=
=
=
☺5
6 a2 – 8 ab – 3a + 4b
(6a2 – 8 ab) – (3a – 4b)
2a (3a – 4b) – 1(3a – 4b)
(3a – 4b)(2a – 1)
(5a2 - 4ab) + (12b3 - 15ab2) Grouping
a(5a - 4b) + 3b2(4b – 5a) Notice the terms are reversed so we need
a(5a - 4b) – 3b2(5a – 4b) to take out a negative to reverse them.
(5a – 4b)(a - 3b2)
Exercises:
1. 6mx - 4m + 3rx - 2r
2. 3my - ab + am - 3by
(reorder like ex 4)
3. a2 - 2ab + a - 2b
4. 3m2 - 5m2p + 3p2 - 5p3
5. 5a2 - 4ab + 12b3 - 15ab2
6. 4ax - 14bx + 35by - 10ay
7. 6a2 - 6ab + 3cb - 3ca (take out CM (3) first)
8. ax + 2a2x - a - 2a2 (take out CM (a) first)
9. a3 - a2b + ab2 - b3
10. 2x3 - 5xy2 - 2x2y + 5y3
11. x2 - 8x + 7x - 56
12. a2 + 8ab + 7ab + 56b2
13. c2 - 8cd - 7cd + 56d2
14. 6x2 - 4x - 9x + 6
15. 6x2 + 9xy -4xy -6y2
16. 24a2 - 9ab + 32ab - 12b2
17. 6x2 -8x -9x + 12
18. 12x2 + 3x - 20x - 5
19. 2x2 + 8x - 3x - 12
20. x2 + 3x - 5x - 15
21. a2x2 + abx2 + 3a2x + 3abx
22. 2abc + 3b2c – 2ac2 – 3bc2
23. 6x2 – 6xy + 3yz – 3xz
24. 6p2 + 9pt – 4 pt – 6t2
Method 3- Regular Trinomial factoring form: ax2 + bx + c
Let’s start by reviewing the FOIL method of multiplication of binomials.
1. (x + 8)(x + 3) = x2 + 3x + 8x+ 24 = x2 + 11x + 24
4
(Notice 24 is the PRODUCT of 8 and 3 and 11 is the SUM of 8 and 3)
2. (x + 2)(x + 12) = x2 + 12x + 2x+ 24 = x2 + 14x + 24
(Notice 24 is the PRODUCT of 2 and 12 and 14 is the SUM of 2 and 12)
3. (x + 4)( x – 6) = x2 - 6x + 4x - 24 = x2 -2x - 24
(Notice -24 is the PRODUCT of 4 and -6 and 2 is the SUM of 4 and -6)
Now let’s reverse this process:
RTF
First Type a = 1
i.e.
x2 + bx + c
Regular Trinomial Factorization: RTF is used when a trinomial expression can be factored and the leading
coefficient is 1. To accomplish this factorization first put the expression in descending order and factor out
any CM factors. Then find two factors of the last term that add to be the middle term. These two numbers
can be used as the coefficients of the second terms of the two binomial factors.
If there are no numbers that multiply to be c and add to be b, the trinomial is PRIME.
1. C is positive:
Ex 1: x2 + 6x + 8
The factors of 8 are
±1 and ±8, ±2 and ±4.
Only +2 and +4 have a sum of +6. Therefore
x2 + 6x + 8 = (x + 2)(x + 6). Note: If c is positive and b is positive, the 2 factors are both positive.
Check by FOILing.
Ex 2: x2 - 9x + 8
The factors of 8 are
±1 and ±8, ±2 and ±4.
Only -1 and -8 have a sum of -9. Therefore
2
x – 9x + 8 = (x - 1)(x - 8). Note: If c is positive and b is negative, the 2 factors are both negative.
Check by FOILing.
☺6
Examples:
1. x2 - 11x + 24
2. x2 + 13x + 36
3. x2 + 9x + 14
4. x2 - 5x + 4
5. x2 + 22x + 21
6. 36 - 13x + x2 (hint: put in decreasing order first)
7. x2 - 6x + 5
8. a2 + 2a + 1
9. x2 + 5x + 6
10. x2 + 12x + 35
11. x2 + 7x + 12
12. x2 + 12x + 27
13. x2 + 8x + 15
14. x2 - 8x + 15
15. x2 + 10x + 21
16. x2 + 22x + 21
2. C is negative
Ex. 1
x2 + x – 12
The factors of -12 are
-1 and 12,
1 and -12,
-2 and 6,
2 and -6,
-3 and 4,
3 and -4
2
Only -3 and 4 have a sum of 1. Therefore x + x – 12 = (x - 3)(x + 4). Note: If c is negative, the 2
factors are opposites, with the larger having the positive sign of the middle term. Check by FOILing.
Ex. 2
x2 - x – 12
The factors of -12 are
-1 and 12,
1 and -12,
-2 and 6,
2 and -6,
-3 and 4,
3 and -4
Only 3 and -4 have a sum of -1. Therefore x2 + x – 12 = (x - 4)(x + 3). Note: If c is negative, the 2
factors are opposites, with the larger having the negative sign of the middle term. Check by FOILing
Examples:
☺7
1. x2 - 5x - 24
2. x2 + 5x – 36
3. x2 + 13x - 14
4. x2 - 5x – 24
5. x2 - 4x - 21
6. x2 - 2x - 15
7. x2 - 4x - 5
8. a2 + 2a - 3
5
9. x2 + 5x – 8
10. x2 + 6x – 27
11. x2 + 2x – 15
12. x2 - 14x – 15
Now let’s combine the two types, with a +c or –c.
1. x2 + 20x + 64
2. a2 - 22a + 121
3. x2 + 2x – 8
4. x2 + 6x – 27
5. x2 + 4x - 5
6. x2 + 12x + 27
7. x2 + 3x – 27
8. x2 + 2x – 24
9. x2 + 3x – 18
10. x2 + 19x + 18
In the following remove a common factor, then use trinomial factoring.(if possible)
1. 3x2 + 18x - 81
2. 2a2 - 44a + 242
3. 5x2 + 10x – 40
4. 2x2 + 12x – 54
5. 6x2 + 24x - 30
6. abx2 + 12abx +27ab
7. bx2 + 3bx – 27b
8. -x2 - 2x + 24 (hint: remove -1)
9. -3ax2 - 9ax + 54a (hint: remove -3a)
10. a2x2 + 19a2x + 18a2
Before we go to the next type, let’s try a mixture of the first 3 types.
Remember:
(1) Look for a common factor and remove it
(2) If there are 4 terms, try factoring by grouping
(3) If there are 3 terms, try regular trinomial factoring
Examples:
☺8
1. 2a3 - 2a2b + 2ab2 - 2b3
2. 3x2 + 6x – 24
3. 14xyz - 18xyz2
4. 36 + 13x + x2
5. 6ax2 – 4ax – 9ax + 6a
6. x2 + 18x + 81
7. x2 + 3x – 18
8. 12ax + bx + 32cx
9. 2a2b - 44ab + 242 b
10. 2x2 + 8x - 3x - 12
11. a2 - 24a + 144
12. c4 + c6 - c2
13. x2 + x – 12
14. x2 + 13x + 12
15. 3my - ab + am - 3by
16. x2 + 25x – 54
(Try these) Notice the product is an odd number and the sum is zero (x2 + 0x – c)
17. x2 – 25
18. x2 – 144
19. x2 – 1
20. x2 – 49
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Method 3- Regular Trinomial factoring form: ax2 + bx + c
Let’s start by reviewing the FOIL method of multiplication of binomials where a ≠ 1.
1.(2x + 5)(x + 3) = 2x2 + 6x + 5x+ 15 = 2x2 + 11x + 15
(Notice 15 is the PRODUCT of 5 and 3, and 2x2 is the PRODUCT of 2x and x, and
11x is the SUM of 6x and 5x)
2.(3x + 2)(x + 12) = 3x2 + 36x + 2x + 24 = 3x2 + 38x + 24
(Notice 24 is the PRODUCT of 2 and 12, and 3x2 is the PRODUCT of 3x and x, and
38x is the SUM of 36x and 2x)
3.(5x + 4)( x – 6) = 5x2 - 30x + 4x - 24 = 5x2 - 26x - 24
(Notice -24 is the PRODUCT of 4 and -6 and 5x2 is the PRODUCT of 5x and x, and
-26x is the SUM of -30x and 4x)
Now let’s reverse this process:
Second Type
a≠1
i.e.
ax2 + bx + c
This can be done by trial and error (the old way), factoring by grouping, or by a unique short cut
that cuts computation drastically.
Ex 1: 2x2 + 11x + 15
Step 1: Multiply a and c to get the product (2 • 15 =30)
Step 2: Repeat the process from the first method (where a = 1) and find the factors of 30 that have a sum
of b, or 11. This would be 6 and 5.
Step 3: Place the 2 factors under the first term (without the square) and reduce. This gives the factors.
2x
x
2x
2x
i.e.
and
and the two factors are (x + 3)(2x + 5). If you FOIL this to check,


6
3
5
5
you can verify your answer.
Ex 2: 3x2 + 38x + 24
Step 1: Multiply a and c to get the product (3• 24 = 72)
Step 2: Find the factors of 72 that have a sum of b, or 38. This would be 2 and 36.
Step 3: Place the 2 factors under the first term (without the square) and reduce. This gives the factors.
3x
3x
2x
x
i.e.
and
and the two factors are (3x + 2)(x + 18).


2
2
36 18
If you FOIL this to check, you can verify your answer.
Ex 3: 5x2 - 26x – 24
Step 1: Multiply a and c to get the product (5 • -24 = -120)
Step 2: Find the factors of -120 that have a sum of b, or -26. This would be -30 and 4.
Step 3: Place the 2 factors under the first term (without the square) and reduce. This gives the factors.
5x
5x
5x
x
i.e.
and
and the two factors are (5x + 4)(x - 6).


4
4
30 6
If you FOIL this to check, you can verify your answer.
7
☺9
Now try these for yourself: (Remember to factor out a common term first, if necessary)
1. 4x2 + 11x + 6
2. 3x2 – 2x – 21
3. 6x2 + 16x + 10 (hint: factor out 2)
4. 7x2 + 22x + 3
5. 32x2 – 40x + 12(hint: factor out 4)
6. 6x2 – 11x + 4
7. 2x2 + 3x – 14
8. 2x2 – 5x – 12
9. 2x2 – 9x – 18
10. 3a2 – 10a – 8
11. 10 + 19x + 6x2
12. 3x2 + 6x – 9 (hint: CM)
13. 15x2 – 13x + 2
14. 12x2 – 18x + 3
15. 9j2 + 30j + 25
16. 10a2 – 37a + 27
17. 5x2 + 27x + 10
18. 10n2 – 11n – 6
19. 2a2 – 9a – 18
20. 5d2 + 6d – 8
21. 6r2 – 14r + 12
22. -3t2 + 14t -16 (hint: factor out -1)
23. 8a2 – 17a + 9
24. 21d2 – 15 d – 6
8
Method 4 - (Special Case) - Difference of Squares
i.e. a2 – b2 = (a - b)(a + b)
This is probably the easiest form of factoring, used when you have only 2 terms. Remember the special type of
problem earlier where there were only 2 terms, each of which was a perfect square.
Ex. 1:
x2 – 25 = (x + 5)(x – 5) because both x2 and 25 are perfect squares and the middle term is zero.
The product of +5 and -5 is -25 and their sum is 0.
Ex 2:
x2 – 144 = (x + 12)(x – 12) because both x2 and 144 are perfect squares and the middle term is
zero. The product of +12 and -12 is -144 and their sum is 0.
Ex 3:
x2 – 1 = (x + 1)(x – 1)
Ex 4:
x2 – 49 = (x + 7)(x – 7)
Ex 5:
9x2 – 49y2 = (3x)2 – (7y)2 = (3x + 7y)(3x – 7y)
Ex 6:
4x2 – 9y2z2 .
First, there is no CM possible.
Since this can be expressed as (2x)2 – (3yz)2, the answer is (2x –3yz)(2x + 3yz)
Ex. 7:
98x – 338x3
Use CM first. 98x – 338x3 = 2x (49 – 169x2) = 2x [ (7)2 – (13x)2 ] = 2x (7 + 13x)(7 – 13x)
☺10
Examples: (remember to check for a CM first, then look for difference of squares)
1. a2 - 9
2. 4x2 - 9y2
3. 1 - 9d2
4. 49 - c2d2
5. a2 – 25
6. 2z2 - 98
7. 49x2 - 256
8. 45x2 - 20y2z2
9. 0.01n2 - 1.69
a2 
b2
10. -16 + 49x2
11.
13. 121 – 16p2
14. 64 – 16p2
9
4
49
25
12. 9x4 - 16y2
15.
n2
4
 16
9
Method 5 - (Special Case) - Perfect Square
i.e. a2 + 2ab + b2 = (a + b)(a + b)
Let’s start by reviewing the FOIL method of multiplication of binomials like (a + b) 2.
(2x + 5) 2 = (2x + 5) (2x + 5) = 4x2 + 10x + 10x+ 25 = 4x2 + 20x + 25
(Notice 25 is the PRODUCT of 5 and 5, and 4x2 is the PRODUCT of 2x and 2x, and
20x is the SUM of 10x and 10x)
(2x + 5)2 can be factored (2x) 2 + 2•(2x)•(5) + 52 if you recognize it as a perfect square.
A perfect square can be recognized by the following:
4x2 + 20x + 25
1. First term must
be a perfect square
4x2 = (2x)2
3. Middle term must be
twice the product of the
square roots of the 1st and
3rd terms
2. Last term must
be a perfect square
25 = 52
The factorization of a perfect square trinomial is always the (square root of the first term, sign of the middle
term, square root of the last term)2.
Ex 1:
9x2 – 12x + 4
= (3x)2 – 2•(3x)•(2) + 22 = (3x – 2)2
Ex 2:
4y2 – 44y + 121 = (2y)2 – 2•(2y)•(11) + (11)2 = (2y - 11)2
Ex 3:
9x3 + 42x2 + 49x = x(9x2 + 42x + 49) = x[ (3x)2 + 2•(3x)•(7) + 72 ] = x(3x + 7)2
☺11
Exercises:
1. x2 + 16x + 64
2. a2 - 22a + 121
3. 100x2 + 20x + 1
4. 4 - 28r + 49r2
5. 50x2 - 40x + 8
6. 32 - 48a + 18a2
7. 9 + 102g + 289g2
8. 196x2 - 196x + 49
9. 256m2 - 160m + 25
10. 4y2 - 76y + 361
11. 900x2 + 420x + 49
12. 9a2 - 420ab + 4900b2
13.
m2
4
25
 54 m  16
14.
49
9
9
g 2  gh  196
h2
15. 256x2 -288x + 81
*Special case for Trinomial factoring: There may be cases in the 2nd case of trinomial factoring
(ax2 + bx + c, a ≠ 1) where you have the third term with variables also. You then make a small adjustment
to the short-cut method.
Ex: 5x2 – 21xy + 4y2
Notice that the 1st and 3rd terms have variables. The PRODUCT is 5 • 4 or 20. The SUM is -21, Therefore the
factors are -20 and -1. When you place the 2 factors under the first term (without the square), also include the
variable from the 3rd term in the denominator. Then reduce. This gives the factors.
5x
5x
x
i.e.
=
and
, and the two factors are (x – 4y)(5x – y). Check by FOIL.
 20 y  4 y
 1y
10
Completing the Square
This is a useful strategy to employ when you have a perfect square trinomial
that is set up equal to a constant.
x2 + 10x + 25 = 49
For example . . .
If this were only x2 + 10x + 25 . . . it would factor as a sum of squares (x + 5)2
However, we cannot ignore the “ = 49” . . .
Here are the steps to solving by completing the square:
1 – Original equation
2 – Factor the perfect square trinomial
x2 + 10x + 25 = 49
(x + 5)2 = 49
3 – Square root property
x + 5 = + √49
4 – Simplify the radical
x+5=+7
5 – Subtract 5 from both sides, to isolate x
x = -5 + 7
6 – Write as TWO equations
x=-5+7
7 – Solve each equation for x
x=2
8 – Your two solutions (roots) are
x = 2, x = -12
x = -5 – 7
x = -12
9 – Plug each solution in to the original equation to check your solutions.
See Lesson 5-5 for more examples.
☺12
Now try these for yourself: (Remember to factor out a common term first, if necessary)
X2 + 4x + 4 = 25
X2 – 10x + 25 = 49
X2 + 8x + 16 = 7
X2 – 6x + 9 = 8
X2 – 8x + 15 = 0
X2 – 10X + 28 = 0
11
Factoring Worksheet
Now to review all types of factoring:
Remember the four steps:
Step 1: Use CM to remove any common factor
Step 2: If you have two terms, look for Difference of squares
Step 3: If you have 4 terms, try Factoring by grouping
Step 4: If you have 3 terms, first check for a Perfect Square; if it is not a perfect square, use Trinomial
factoring. . If you have a variable in the 3rd term remember the *Special case for Trinomial factoring.
Also: Remember to keep factoring if necessary. i.e.. if you still have difference of squares**
Factor Completely. If it will not factor, indicate PRIME
1.
27x2z – 36xyz + 12y2z
3.
4x2 + 40x + 100
5.
x3y – 11x2y + 28xy
6.
x2y2 - 64y2
7.
3x2y2 - 48y2
8.
2abc + 3b2c – 2ac2 – 3bc2
9.
–3x3 – 30x2 – 72x
10.
8a2 + 56a + 98
5x3y + 21x2y2 + 4xy3
12.
– 72y2 + 8z2
*11.
2.
*4.
3x2 – 108
40x2 – 50xy – 15y2
13.
32a4b2 – 8a2b2
15.
35x2 – 14x + 84x3
16.
16x2 – 60x – 54
17.
15x2y – 75xy + 45xy2
18.
36ax + 3ax2 – 3ax3
19.
a2x2 + abx2 + 3a2x + 3abx
20.
y3 - 121x3y3
21.
d2 + 5d – 9b4 – 81a4b4
23.
648x4y + 375xy4
24.
108x4y2 – 3x2y2
25.
54a2z – 6az – 4z
26.
16x2 – 49y2
27.
y5 – 100y3
28.
x + x2y + x3y3
29.
2a3 - 2a2b + 2ab2 - 2b3
30.
9x2 + 30xy + 25y2
31.
25x2 + 10x + 1
**32. 4hk2 – 4k3 – 9h3 + 9h2k
**33.
x4 – 10x2 + 24
**34.
**35.
b4 – 81a4b4
*14.
*22.
☺13
48x3y – 72x2y2 + 27xy3
5a3b2c – 37a2b2c2 – 72ab2c3
3x3 – 9x2 – 12x + 36
12
Applications of factoring
Factoring is used to find the solutions of polynomials of degree higher than one.
Using the zero-product property, if a • b = 0, then at least one term, a or b, must = 0.
Ex. 1 (x – 1)( x + 2) = 0
Therefore x – 1 = 0 or x + 2 = 0
If x – 1 = 0, then x = 1. If x + 2 = 0, then x = -2.
Ex 2: a2 – 9 = 0
Therefore (a – 3)(a + 3) = 0
If a - 3 = 0, then a = 3. If a + 3 = 0, then a = -3.
Ex 3: x2 + x = 12
Therefore x2 + x – 12 = 0
And (x + 4)(x - 3) = 0
If x + 4 = 0, then x = -4. If x - 3 = 0, then x = 3.
Step 1: Set = 0
Step 2: Factor
Step 3: Set each factor = 0
Step 4: Solve each linear equation
Exercises:
☺14
1. x2 - 11x + 24 = 0
2. x2 + 5x = 36
3. 4x2 + 40x + 100 = 0
4. x2 + 64 = -16x
5. x2 + 9x = - 14
6. x2 + 4 = 5x
7. 2x2 + 3x = 14
8. 36 - 13x + x2 = 0
9. x2 - 8x + 7x = 56
10. 25x2 + 10x + 1 = 0
11. -16 + 49x2 = 0
12. y5 – 100y3 = 0
13. -3t2 + 5t – 16 = 0
14. x2 + 25x = 54
(x – 5)2 = 9
Take the square root of both sides:
Special Type:
(x - 5) 2  9
( x  5)  3
x  5  3, or 8 and 2
15. (x – 9) = 16
2
16. (x – 9)2 = 7
17. 25x2 + 10x + 1 = 9 (hint: factor left, a perfect square)
18. 4 - 28r + 49r2 = 5
19. z2 + 14z + 49 = 16
20. 9x2 – 12x + 4 = 5
13
THE DISCRIMINANT
We have seen that the SOLUTIONS or ROOTS of a quadratic equation are the x-intercepts.
Depending on the particular quadratic equation – its graph will fall into one of the following categories:
1a – It crosses the x-axis twice = it has TWO real, rational roots
1b – It crosses the x-axis twice = it has TWO real irrational roots (non-terminating, non-repeating decimals)
2 – The vertex lies ON the x-axis = it has ONE real, rational root
3a – The vertex lies below the x-axis and the parabola opens DOWN = 2 imaginary roots (never crosses x-axis)
3b – The vertex lies above the x-axis and the parabola opens UP = 2 imaginary roots (never crosses x-axis)
EXAMPLES
1a
1b
2
3a
3b
We can use the DISCRIMINANT to evaluate which of the above categories a quadratic belongs to.
b2 – 4ac
Formula for the DISCRIMINANT
Finding the value of the discriminant tells you how MANY roots and what TYPE of roots there are.
Let’s revisit the above categories, in light of the value of the discriminant.
Depending on the particular quadratic equation – its graph will fall into one of the following categories:
1a – It crosses the x-axis twice = it has TWO real, rational roots
if b2 – 4ac = a perfect square
1b – It crosses the x-axis twice = it has TWO real irrational roots if b2 – 4ac = NOT a perfect square
2 – The vertex lies ON the x-axis = it has ONE real, rational root
if b2 – 4ac = ZERO
3a – Vertex lies below x-axis and the parabola opens DOWN = 2 imaginary roots if b2 – 4ac < ZERO
3b – Vertex lies above x-axis and the parabola opens UP = 2 imaginary roots if b2 – 4ac < ZERO
14
☺15
Find the value of the discriminant in the following quadratic equations. Based on that
information, tell how many and what kind of root(s) it has.
Quadratic
Discriminant
# roots
type of root(s)
15
QUADRATIC FORMULA
We have discussed several methods for SOLVING – or finding the ROOTS – of a quadratic.
There is one, final method that is particularly useful when other methods are not – specifically,
when your coefficients are fractions or decimals, etc. This is the QUADRATIC FORMULA.
It is a combination of the vertex formula and the discriminant.
x
=
- b + √b2-4ac
2a
☺16
Solve the following quadratics by using the quadratic formula.
16
☺17 – FINAL STEP!
A1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
B1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
C1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A
B
C
Write in standard form.
A=
B=
C=
Axis of Symmetry =
Vertex =
Opens =
Y intercept =
Determinant =
How many roots?
What type?
Factor by any method (choose carefully)
What are the roots?
Graph and label x-intercept, y-intercept, vertex, and axis of symmetry.
Shade with HORIZONTAL stripes where y > the quadratic.
Shade with VERTICAL stripes where y < the quadratic.
Label the x axis and y axis
Use and label consistent unit on each axis
Give a table of values for x and y (at least 5 ordered pairs).
Is it a relation? Is it a function? Is it linear? Is it quadratic?
Write in standard form.
A=
B=
C=
Axis of Symmetry =
Vertex =
Opens =
Y intercept =
Determinant =
How many roots?
What type?
Factor by any method (choose carefully)
What are the roots?
Graph and label x-intercept, y-intercept, vertex, and axis of symmetry.
Shade with HORIZONTAL stripes where y > the quadratic.
Shade with VERTICAL stripes where y < the quadratic.
Label the x axis and y axis
Use and label consistent unit on each axis
Give a table of values for x and y (at least 5 ordered pairs).
Is it a relation? Is it a function? Is it linear? Is it quadratic?
Write in standard form.
A=
B=
C=
Axis of Symmetry =
Vertex =
Opens =
Y intercept =
Determinant =
How many roots?
What type?
Factor by any method (choose carefully)
What are the roots?
Graph and label x-intercept, y-intercept, vertex, and axis of symmetry.
Shade with HORIZONTAL stripes where y > the quadratic.
Shade with VERTICAL stripes where y < the quadratic.
Label the x axis and y axis
Use and label consistent unit on each axis
Give a table of values for x and y (at least 5 ordered pairs).
Is it a relation? Is it a function? Is it linear? Is it quadratic?
17