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Adding and Subtracting Polynomials
Lesson 9-1
Algebra 1
Adding and Subtracting Polynomials
Lesson 9-1
Algebra 1
Find the degree of each monomial.
a. 18
Degree: 0
The degree of a nonzero constant is 0.
b. 3xy3
Degree: 4
The exponents are 1 and 3. Their sum is 4.
c. 6c
Degree: 1
6c = 6c1. The exponent is 1.
Adding and Subtracting Polynomials
Lesson 9-1
Algebra 1
Write each polynomial in standard form. Then name each
polynomial by its degree and the number of its terms.
a. –2 + 7x
7x – 2
Place terms in order.
linear binomial
b. 3x5 – 2 – 2x5 + 7x
3x5 – 2x5 + 7x – 2
Place terms in order.
x5 + 7x – 2
Combine like terms.
fifth degree trinomial
Adding and Subtracting Polynomials
Lesson 9-1
Algebra 1
Simplify (6x2 + 3x + 7) + (2x2 – 6x – 4).
Method 1: Add vertically.
Line up like terms. Then add the coefficients.
6x2 + 3x + 7
2x2 – 6x – 4
8x2 – 3x + 3
Method 2: Add horizontally.
Group like terms. Then add the coefficients.
(6x2 + 3x + 7) + (2x2 – 6x – 4) = (6x2 + 2x2) + (3x – 6x) + (7 – 4)
= 8x2 – 3x + 3
Adding and Subtracting Polynomials
Lesson 9-1
Algebra 1
Simplify (2x3 + 4x2 – 6) – (5x3 + 2x – 2).
Method 1: Subtract vertically.
Line up like terms. Then add the coefficients.
(2x3 + 4x2
–(5x3
– 2x
– 6)
– 2)
Line up like terms.
2x3 + 4x2
–5x3
– 2x
–3x3 + 4x2 – 2x
–6
+2
–4
Add the opposite.
Adding and Subtracting Polynomials
Lesson 9-1
Algebra 1
(continued)
Method 2: Subtract horizontally.
(2x3 + 4x2 – 6) – (5x3 + 2x – 2)
= 2x3 + 4x2 – 6 – 5x3 – 2x + 2
Write the opposite of each term in
the polynomial being subtracted.
= (2x3 – 5x3) + 4x2 – 2x + (–6 + 2)
Group like terms.
= –3x3 + 4x2 – 2x – 4
Simplify.
Multiplying and Factoring
Lesson 9-2
Algebra 1
Multiplying and Factoring
Lesson 9-2
Algebra 1
Simplify –2g2(3g3 + 6g – 5).
–2g2(3g3 + 6g – 5)
= –2g2(3g3) –2g2(6g) –2g2(–5)
Use the Distributive Property.
= –6g2 + 3 – 12g2 + 1 + 10g2
Multiply the coefficients and add the
exponents of powers with the same base.
= –6g5 – 12g3 + 10g2
Simplify.
Multiplying and Factoring
Lesson 9-2
Algebra 1
Find the GCF of 2x4 + 10x2 – 6x.
List the prime factors of each term. Identify the factors common to all terms.
2x4 = 2 • x • x • x • x
10x2 = 2 • 5 • x • x
6x = 2 • 3 • x
The GCF is 2 • x, or 2x.
Multiplying and Factoring
Lesson 9-2
Algebra 1
Factor 4x3 – 8x2 + 12x.
Step 1: Find the GCF.
=2•2•x•x•x
8x2 = 2 • 2 • 2 • x • x
12x = 2 • 2 • 3 • x
4x3
The GCF is 2 • 2 • x, or 4x.
Step 2: Factor out the GCF.
4x3 – 8x2 + 12x
= 4x(x2) + 4x(–2x) + 4x(3)
= 4x(x2 – 2x + 3)
Multiplying Binomials
Lesson 9-3
Algebra 1
Multiplying Binomials
Lesson 9-3
Algebra 1
Simplify (2y – 3)(y + 2).
(2y – 3)(y + 2) = (2y – 3)(y) + (2y – 3)(2)
Distribute 2y – 3.
= 2y2 – 3y + 4y – 6
Now distribute y and 2.
= 2y2 + y – 6
Simplify.
Multiplying Binomials
Lesson 9-3
Algebra 1
Simplify (4x + 2)(3x – 6).
First
(4x + 2)(3x – 6)
Outer
Inner
Last
= (4x)(3x) + (4x)(–6) + (2)(3x) + (2)(–6)
= 12x2
–
= 12x2
–
The product is 12x2 – 18x – 12.
24x
+
18x
6x
–
12
–
12
Multiplying Binomials
Lesson 9-3
Algebra 1
Find the area of the shaded region. Simplify.
area of outer rectangle = (3x + 2)(2x – 1)
area of hole = x(x + 3)
area of shaded region = area of outer rectangle – area of hole
= (3x + 2)(2x – 1)
–x(x + 3)
Substitute.
= 6x2 – 3x + 4x – 2
–x2 – 3x
Use FOIL to simplify (3x + 2) (2x – 1) and
the Distributive Property to simplify x(x + 3).
= 6x2 – x2 – 3x + 4x – 3x – 2
Group like terms.
= 5x2 – 2x – 2
Simplify.
Multiplying Binomials
Lesson 9-3
Algebra 1
Simplify the product (3x2 – 2x + 3)(2x + 7).
Method 1: Multiply using the vertical method.
3x2 – 2x + 3
2x + 7
21x2 – 14x + 21
Multiply by 7.
6x3 – 4x2 + 6x
Multiply by 2x.
6x3 + 17x2 – 8x + 21
Add like terms.
Multiplying Binomials
Lesson 9-3
Algebra 1
(continued)
Method 2: Multiply using the horizontal method.
(2x + 7)(3x2 – 2x + 3)
= (2x)(3x2) – (2x)(2x) + (2x)(3) + (7)(3x2) – (7)(2x) + (7)(3)
= 6x3 – 4x2 + 6x + 21x2 – 14x + 21
= 6x3 + 17x2 – 8x + 21
The product is 6x3 + 17x2 – 8x + 21.
Multiplying Special Cases
Lesson 9-4
Algebra 1
Multiplying Special Cases
Lesson 9-4
Algebra 1
a. Find (y + 11)2.
(y + 11)2 = y2 + 2y(11) + 72
= y2 + 22y + 121
Square the binomial.
Simplify.
b. Find (3w – 6)2.
(3w – 6)2 = (3w)2 –2(3w)(6) + 62
= 9w2 – 36w + 36
Square the binomial.
Simplify.
Multiplying Special Cases
Lesson 9-4
Algebra 1
Among guinea pigs, the black fur gene (B) is dominant and
the white fur gene (W) is recessive. This means that a guinea pig with
at least one dominant gene (BB or BW) will have black fur. A guinea
pig with two recessive genes (WW) will have white fur.
The Punnett square below models the possible combinations of color
genes that parents who carry both genes can pass on to their
offspring. Since WW is 1 of the outcomes, the probability that a guinea
4
pig has white fur is 1 .
4
B
B
W
BB BW
W BW WW
Multiplying Special Cases
Lesson 9-4
Algebra 1
(continued)
You can model the probabilities found in the Punnett square with the
expression ( 1 B + 1 W)2. Show that this product gives the same result
2
2
as the Punnett square.
(12 B + 12 W)2 = ( 12 B)2 – 2(
1
1
1
1
B)( W) +
2
2
1
= 4 B2 + 2 BW + 4 W 2
( 12 W)2
Square the binomial.
Simplify.
The expressions 1 B2 and 1 W 2 indicate the probability that offspring will
4
4
have either two dominant genes or two recessive genes is 1 . The
4
1
1
expression BW indicates that there is chance that the offspring will
2
2
inherit both genes. These are the same probabilities shown in the
Punnett square.
Multiplying Special Cases
Lesson 9-4
Algebra 1
a. Find 812 using mental math.
812 = (80 + 1)2
= 802 + 2(80 • 1) + 12
Square the binomial.
= 6400 + 160 + 1 = 6561
Simplify.
b. Find 592 using mental math.
592 = (60 – 1)2
= 602 – 2(60 • 1) + 12
Square the binomial.
= 3600 – 120 + 1 = 3481
Simplify.
Multiplying Special Cases
Lesson 9-4
Algebra 1
Find (p4 – 8)(p4 + 8).
(p4 – 8)(p4 + 8) = (p4)2 – (8)2
= p8 – 64
Find the difference of squares.
Simplify.
Multiplying Special Cases
Lesson 9-4
Algebra 1
Find 43 • 37.
43 • 37 = (40 + 3)(40 – 3)
Express each factor using 40 and 3.
= 402 – 32
Find the difference of squares.
= 1600 – 9 = 1591
Simplify.
Factoring Trinomials of the Type x2 + bx + c
Lesson 9-5
Algebra 1
Factoring Trinomials of the Type x2 + bx + c
Lesson 9-5
Algebra 1
Factor x2 + 8x + 15.
Find the factors of 15. Identify the pair that has a sum of 8.
Factors of 15
1 and 15
3 and 5
Sum of Factors
16
8
x2 + 8x + 15 = (x + 3)(x + 5).
Check: x2 + 8x + 15
(x + 3)(x + 5)
= x2 + 5x + 3x + 15
= x2 + 8x + 15
Factoring Trinomials of the Type x2 + bx + c
Lesson 9-5
Algebra 1
Factor c2 – 9c + 20.
Since the middle term is negative, find the negative factors of 20.
Identify the pair that has a sum of –9.
Factors of 20
–1 and –20
–2 and –10
–4 and –5
Sum of Factors
–21
–12
–9
c2 – 9c + 20 = (c – 5)(c – 4)
Factoring Trinomials of the Type x2 + bx + c
Lesson 9-5
Algebra 1
a. Factor x2 + 13x – 48.
Identify the pair of factors of –48
that has a sum of 13.
Factors of –48
1 and –48
48 and
–1
2 and –24
24 and
–2
3 and –16
16 and
–3
Sum of Factors
–47
47
–22
22
–13
13
x2 + 13x – 48 = (x + 16)(x – 3)
b. Factor n2 – 5n – 24.
Identify the pair of factors of –24 that
has a sum of –5.
Factors of –24
1 and –24
24 and –1
2 and –12
12 and –2
3 and –8
Sum of Factors
–23
23
–10
10
–5
n2 – 5n – 24 = (n + 3)(n – 8)
Factoring Trinomials of the Type x2 + bx + c
Lesson 9-5
Algebra 1
Factor d2 + 17dg – 60g2.
Find the factors of –60.
Identify the pair that has a sum of 17.
Factors of –60
1 and –60
60 and
–1
2 and –30
30 and
–2
3 and –20
20 and
–3
Sum of Factors
–59
59
–28
28
–17
17
d2 + 17dg – 60g2 = (d – 3g)(d + 20g)
Factoring Trinomials of the Type ax2 + bx + c
Lesson 9-6
Algebra 1
Factoring Trinomials of the Type ax2 + bx + c
Lesson 9-6
Algebra 1
Factor 20x2 + 17x + 3.
20x2
F
factors
of a
O
+ 17x
+3
I
L
1 • 20
1 • 3 + 1 • 20 = 23
1 • 1 + 3 • 20 = 61
1•3
3•1
2 • 10
2 • 3 + 1 • 10 = 16
2 • 1 + 3 • 10 = 32
1•3
3•1
4•5
4 • 3 + 1 • 5 = 17
1•3
20x2 + 17x + 3 = (4x + 1)(5x + 3)
factors
of c
Factoring Trinomials of the Type ax2 + bx + c
Lesson 9-6
Algebra 1
Factor 3n2 – 7n – 6.
3n2
(1)(3)
–7n
–6
(1)(–6) + (1)(3) = –3
(1)(–6)
(1)(1) + (–6)(3) = –17
(–6)(1)
(1)(–3) + (2)(3) = 3
(2)(–3)
(1)(2) + (–3)(3) = –7
(–3)(2)
3n2 – 7n – 6 = (n – 3)(3n + 2)
Factoring Trinomials of the Type ax2 + bx + c
Lesson 9-6
Algebra 1
Factor 18x2 + 33x – 30 completely.
18x2 + 33x – 30 = 3(6x2 + 11x – 10)
Factor out the GCF.
Factor 6x2 + 11x – 10.
6x2
(2)(3)
+ 11x
(2)(–10) + (1)(3) = –17
(2)(1) + (–10)(3) = –28
(2)(–5) + (2)(3) = –4
(2)(2) + (–5)(3) = –11
(2)(–2) + (5)(3) = 11
–10
(1)(–10)
(–10)(1)
(2)(–5)
(–5)(2)
(5)(–2)
6x2 + 11x – 10 = (2x + 5)(3x – 2)
18x2 + 33x – 30 = 3(2x + 5)(3x – 2)
Include the GCF in your final answer.
Factoring Special Cases
Lesson 9-7
Algebra 1
Factoring Special Cases
Lesson 9-7
Algebra 1
Factor m2 – 6m + 9.
m2 – 6m + 9 = m • m – 6m + 3 • 3
= m • m – 2(m • 3) + 3 • 3
= (m – 3)2
Rewrite first and last terms.
Does the middle term equal 2ab?
6m = 2(m • 3)
Write the factors as the square
of a binomial.
Factoring Special Cases
Lesson 9-7
Algebra 1
The area of a square is (16h2 + 40h + 25) in.2 Find the
length of a side.
16h2 + 40h + 25 = (4h)2 + 40h + 52
Write 16h2 as (4h)2
and 25 as 52.
= (4h)2 + 2(4h)(5) + 52
Does the middle term equal 2ab? 40h = 2(4h)(5)
= (4h + 5)2
Write the factors as the square of a binomial.
The side of the square has a length of (4h + 5) in.
Factoring Special Cases
Lesson 9-7
Algebra 1
Factor a2 – 16.
a2 – 16 = a2 – 42
= (a + 4)(a – 4)
Check: Use FOIL to multiply.
(a + 4)(a – 4)
a2 – 4a + 4a – 16
a2 – 16
Rewrite 16 as 42.
Factor.
Factoring Special Cases
Lesson 9-7
Algebra 1
Factor 9b2 – 225.
9b2 – 225 = (3b)2 – 152
Rewrite 9b2 as (3b)2 and 225 as 152.
= (3b + 15)(3b –15) Factor.
Factoring Special Cases
Lesson 9-7
Algebra 1
Factor 5x2 – 80.
5x2 – 80 = 5(x2 – 16)
Factor out the GCF of 5.
= 5(x + 4)(x – 4) Factor (x2 – 16).
Check: Use FOIL to multiply the binomials. Then multiply by the GCF.
5(x + 4)(x – 4)
5(x2 – 16)
5x2 – 80
Factoring by Grouping
Lesson 9-8
Algebra 1
Factoring by Grouping
Lesson 9-8
Algebra 1
Factor 6x3 + 3x2 – 4x – 2.
6x3 + 3x2 – 4x – 2 = 3x2(2x + 1) – 2(2x + 1)
= (2x + 1)(3x2 – 2)
Check: 6x3 + 3x2 – 4x – 2
Factor the GCF from each
group of two terms.
Factor out (2x + 1).
(2x + 1)(3x2 – 2)
= 6x3 – 4x + 3x2 – 2
Use FOIL.
= 6x3 + 3x2 – 4x – 2
Write in standard form.
Factoring by Grouping
Lesson 9-8
Algebra 1
Factor 8t4 + 12t3 + 16t2 + 24t.
8t4 + 12t3 + 16t2 + 24t = 4t(2t3 + 3t2 + 4t + 6)
Factor out the GCF, 4.
= 4t[t2(2t + 3) + 2(2t + 3)]
Factor by grouping.
= 4t(2t + 3)(t2 + 2)
Factor again.
Factoring by Grouping
Lesson 9-8
Algebra 1
Factor 24h2 + 10h – 6.
Step 1: 24h2 + 10h – 6 = 2(12h2 + 5h – 3)
Factor out the GCF, 2.
Step 2: 12 • –3 = –36
Find the product ac.
Step 3:
Factors
–2(18) = –36
–3(12) = –36
–4(9) = –36
Sum
–2 + 18 = 16
–3 + 12 = 9
–4 + 9 = 5
Find two factors of ac that
have a sum b. Use mental
math to determine a good
place to start.
Step 4: 12h2 – 4h + 9h – 3
Rewrite the trinomial.
Step 5: 4h(3h – 1) + 3(3h – 1)
Factor by grouping.
(4h + 3)(3h – 1)
24h2 + 10h – 6 = 2(4h + 3)(3h – 1)
Factor again.
Include the GCF in your
final answer.
Factoring by Grouping
Lesson 9-8
Algebra 1
A rectangular prism has a volume of 36x3 + 51x2 + 18x.
Factor to find the possible expressions for the length, width, and
height of the prism.
Factor 36x3 + 51x2 + 18x.
Step 1: 3x(12x2 + 17x + 6) Factor out the GCF, 3x.
Step 2: 12 • 6 = 72
Step 3: Factors
4 • 18
6 • 12
8•9
Find the product ac.
Sum
4 + 18 = 22
6 + 12 = 18
8 + 9 = 17
Find two factors of ac that
have sum b. Use mental
math to determine a good
place to start.
Factoring by Grouping
Lesson 9-8
Algebra 1
(continued)
Step 4: 3x(12x2 + 8x + 9x + 6)
Rewrite the trinomial.
Step 5: 3x[4x(3x + 2) + 3(3x + 2)]
Factor by grouping.
3x(4x + 3)(3x + 2)
Factor again.
The possible dimensions of the prism are 3x, (4x + 3), and (3x + 2).