Algebra 1 Lesson 9-1 Adding and Subtracting Polynomials.

Algebra 1Algebra 1Lesson 9-1

Adding and Subtracting PolynomialsAdding and Subtracting Polynomials

Algebra 1Algebra 1Lesson 9-1


Find the degree of each monomial.

a. 18

b. 3xy3

c. 6c

The degree of a nonzero constant is 0.Degree: 0

The exponents are 1 and 3. Their sum is 4.Degree: 4

6c = 6c1. The exponent is 1.Degree: 1

Algebra 1Algebra 1


Lesson 9-1

Write each polynomial in standard form. Then name each

polynomial by its degree and the number of its terms.

a. –2 + 7x

linear binomial

b. 3x5 – 2 – 2x5 + 7x

x5 + 7x – 2 Combine like terms.

fifth degree trinomial

Place terms in order.3x5 – 2x5 + 7x – 2

Place terms in order.7x – 2

Algebra 1Algebra 1


Lesson 9-1

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

Algebra 1Algebra 1


Simplify (2x3 + 4x2 – 6) – (5x3 + 2x – 2).

Lesson 9-1

Method 1: Subtract vertically.

Line up like terms. Then add the coefficients.

(2x3 + 4x2 – 6) Line up like terms.

–(5x3 – 2x – 2)

2x3 + 4x2 – 6 Add the opposite.

–5x3 – 2x + 2

–3x3 + 4x2 – 2x – 4

Algebra 1Algebra 1


(continued)

Lesson 9-1

Method 2: Subtract horizontally.

= 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.

(2x3 + 4x2 – 6) – (5x3 + 2x – 2)

Algebra 1Algebra 1

Multiplying and FactoringMultiplying and Factoring

Lesson 9-2

Algebra 1Algebra 1


Lesson 9-2

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.

Algebra 1Algebra 1


Lesson 9-2

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.

Algebra 1Algebra 1


Lesson 9-2

Factor 4x3 – 8x2 + 12x.

Step 1: Find the GCF.

4x3 = 2 • 2 • x • x • x

8x2 = 2 • 2 • 2 • x • x

12x = 2 • 2 • 3 • x

Step 2: Factor out the GCF.

4x3 – 8x2 + 12x

= 4x(x2) + 4x(–2x) + 4x(3)

= 4x(x2 – 2x + 3)

The GCF is 2 • 2 • x, or 4x.

Algebra 1Algebra 1

Multiplying BinomialsMultiplying Binomials

Lesson 9-3

Algebra 1Algebra 1


Lesson 9-3

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.

Algebra 1Algebra 1


Lesson 9-3

Simplify (4x + 2)(3x – 6).

The product is 12x2 – 18x – 12.

Last

(2)(–6)+

Outer

(4x)(–6)+

Inner

(2)(3x)+

24x 6x 12– + –= 12x2

= 12x2 18x– 12–

First

= (4x)(3x)(4x + 2)(3x – 6)

Algebra 1Algebra 1


Lesson 9-3

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 – x2 – 3x + 4x – 3x – 2 Group like terms.

= 5x2 – 2x – 2 Simplify.

= 6x2 – 3x + 4x – 2 –x2 – 3x

Use FOIL to simplify (3x + 2) (2x – 1) and the Distributive Property to simplify x(x + 3).

Algebra 1Algebra 1


Lesson 9-3

Simplify the product (3x2 – 2x + 3)(2x + 7).

21x2 – 14x + 21 Multiply by 7.

6x3 – 4x2 + 6x Multiply by 2x.

Method 1: Multiply using the vertical method.

3x2 – 2x + 3

2x + 7

6x3 + 17x2 – 8x + 21 Add like terms.

Algebra 1Algebra 1


Lesson 9-3

(continued)

Method 2: Multiply using the horizontal method.

= 6x3 – 4x2 + 6x + 21x2 – 14x + 21

= 6x3 + 17x2 – 8x + 21

The product is 6x3 + 17x2 – 8x + 21.

= (2x)(3x2) – (2x)(2x) + (2x)(3) + (7)(3x2) – (7)(2x) + (7)(3)

(2x + 7)(3x2 – 2x + 3)

Algebra 1Algebra 1

Multiplying Special CasesMultiplying Special Cases

Lesson 9-4

Algebra 1Algebra 1


Lesson 9-4

a. Find (y + 11)2.

(y + 11)2 = y2 + 2y(11) + 72 Square the binomial.

= y2 + 22y + 121 Simplify.

b. Find (3w – 6)2.

(3w – 6)2 = (3w)2 –2(3w)(6) + 62 Square the binomial.

= 9w2 – 36w + 36 Simplify.

Algebra 1Algebra 1


Lesson 9-4

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 of the outcomes, the probability that a guinea pig has white fur is .

BB BW

BW WW

B

W

B W

14

14

Algebra 1Algebra 1


Lesson 9-4

(continued)

You can model the probabilities found in the Punnett square with the expression ( B + W)2. Show that this product gives the same result as the Punnett square.

12

12

( B + W)2 = ( B)2 – 2( B)( W) + ( W)2 Square the binomial.12

12

12

12

12

12

= B2 + BW + W 2 Simplify.14

12

14

The expressions B2 and W 2 indicate the probability that offspring will

have either two dominant genes or two recessive genes is . The

expression BW indicates that there is chance that the offspring will

inherit both genes. These are the same probabilities shown in the

Punnett square.

12

14

14

12

14

Algebra 1Algebra 1


Lesson 9-4

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.

Algebra 1Algebra 1


Lesson 9-4

Find (p4 – 8)(p4 + 8).

(p4 – 8)(p4 + 8) = (p4)2 – (8)2 Find the difference of squares.

= p8 – 64 Simplify.

Algebra 1Algebra 1


Lesson 9-4

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.

Algebra 1Algebra 1

Factoring Trinomials of the Type x2 + bx + cFactoring Trinomials of the Type x2 + bx + c

Lesson 9-5

Algebra 1Algebra 1


Lesson 9-5

Factor x2 + 8x + 15.

Find the factors of 15. Identify the pair that has a sum of 8.

Factors of 15 Sum of Factors

1 and 15 16

3 and 5 8

x2 + 8x + 15 = (x + 3)(x + 5).

= x2 + 5x + 3x + 15

Check: x2 + 8x + 15 (x + 3)(x + 5)

= x2 + 8x + 15

Algebra 1Algebra 1


Lesson 9-5

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.

c2 – 9c + 20 = (c – 5)(c – 4)

Factors of 20 Sum of Factors

–1 and –20 –21

–2 and –10 –12

–4 and –5 –9

Algebra 1Algebra 1


Lesson 9-5

a. Factor x2 + 13x – 48.

Identify the pair of factors of –48 that has a sum of 13.

b. Factor n2 – 5n – 24.

Identify the pair of factors of –24 that has a sum of –5.

x2 + 13x – 48 = (x + 16)(x – 3) n2 – 5n – 24 = (n + 3)(n – 8)

Factors of –48 Sum of Factors

1 and –48 –47

48 and –1 47

2 and –24 –22

24 and –2 22

3 and –16 –13

16 and –3 13


1 and –24 –23

24 and –1 23

2 and –12 –10

12 and –2 10

3 and –8 –5

Algebra 1Algebra 1


Factor d2 + 17dg – 60g2.

Lesson 9-5


1 and –60 –59

60 and –1 59

2 and –30 –28

30 and –2 28

3 and –20 –17

20 and –3 17

Find the factors of –60.

Identify the pair that has a sum of 17.

d2 + 17dg – 60g2 = (d – 3g)(d + 20g)

Algebra 1Algebra 1

Factoring Trinomials of the Type ax2 + bx + cFactoring Trinomials of the Type ax2 + bx + c

Lesson 9-6

Algebra 1Algebra 1


Lesson 9-6

Factor 20x2 + 17x + 3.

20x2 + 17x + 3

F O I L

2 • 10 2 • 3 + 1 • 10 = 16 1 • 32 • 1 + 3 • 10 = 32 3 • 1

1 • 20 1 • 3 + 1 • 20 = 23 1 • 31 • 1 + 3 • 20 = 61 3 • 1

factors of a

factors of c

4 • 5 4 • 3 + 1 • 5 = 17 1 • 3

20x2 + 17x + 3 = (4x + 1)(5x + 3)

Algebra 1Algebra 1


Lesson 9-6

Factor 3n2 – 7n – 6.

3n2 –7n –6

(1)(3) (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)

Algebra 1Algebra 1


Lesson 9-6

Factor 18x2 + 33x – 30 completely.

18x2 + 33x – 30 = 3(6x2 + 11x – 10) Factor out the GCF.

18x2 + 33x – 30 = 3(2x + 5)(3x – 2) Include the GCF in your final answer.

Factor 6x2 + 11x – 10.

6x2 + 11x –10

6x2 + 11x – 10 = (2x + 5)(3x – 2)

(2)(3) (2)(–10) + (1)(3) = –17 (1)(–10)

(2)(1) + (–10)(3) = –28 (–10)(1)

(2)(–5) + (2)(3) = –4 (2)(–5)

(2)(2) + (–5)(3) = –11 (–5)(2)

(2)(–2) + (5)(3) = 11 (5)(–2)

Algebra 1Algebra 1

Factoring Special CasesFactoring Special Cases

Lesson 9-7

Algebra 1Algebra 1


Lesson 9-7

Factor m2 – 6m + 9.

m2 – 6m + 9 = m • m – 6m + 3 • 3 Rewrite first and last terms.

= m • m – 2(m • 3) + 3 • 3 Does the middle term equal 2ab? 6m = 2(m • 3)

= (m – 3)2 Write the factors as the square of a binomial.

Algebra 1Algebra 1


Lesson 9-7

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 + 5)2 Write the factors as the square of a binomial.

The side of the square has a length of (4h + 5) in.

= (4h)2 + 2(4h)(5) + 52 Does the middle term equal 2ab? 40h = 2(4h)(5)

Algebra 1Algebra 1


Lesson 9-7

Factor a2 – 16.

a2 – 16 = a2 – 42 Rewrite 16 as 42.

= (a + 4)(a – 4) Factor.

Check: Use FOIL to multiply.

(a + 4)(a – 4)

a2 – 4a + 4a – 16

a2 – 16

Algebra 1Algebra 1


Factor 9b2 – 225.

Lesson 9-7

9b2 – 225 = (3b)2 – 152 Rewrite 9b2 as (3b)2 and 225 as 152.

= (3b + 15)(3b –15) Factor.

Algebra 1Algebra 1


Lesson 9-7

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

Algebra 1Algebra 1

Factoring by GroupingFactoring by Grouping

Lesson 9-8

Algebra 1Algebra 1


Lesson 9-8

Factor 6x3 + 3x2 – 4x – 2.

6x3 + 3x2 – 4x – 2 = 3x2(2x + 1) – 2(2x + 1) Factor the GCF from each group of two terms.

= (2x + 1)(3x2 – 2) Factor out (2x + 1).

= 6x3 – 4x + 3x2 – 2 Use FOIL.

Check: 6x3 + 3x2 – 4x – 2 (2x + 1)(3x2 – 2)

= 6x3 + 3x2 – 4x – 2 Write in standard form.

Algebra 1Algebra 1


Lesson 9-8

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.

Algebra 1Algebra 1


Factor 24h2 + 10h – 6.

Lesson 9-8

Step 1: 24h2 + 10h – 6 = 2(12h2 + 5h – 3) Factor out the GCF, 2.

Step 2: 12 • –3 = –36 Find the product ac.

Step 4: 12h2 – 4h + 9h – 3 Rewrite the trinomial.

Step 5: 4h(3h – 1) + 3(3h – 1) Factor by grouping.

(4h + 3)(3h – 1) Factor again.

24h2 + 10h – 6 = 2(4h + 3)(3h – 1) Include the GCF in your final answer.

Step 3: Factors Sum–2(18) = –36 –2 + 18 = 16–3(12) = –36 –3 + 12 = 9 –4(9) = –36 –4 + 9 = 5

Find two factors of ac that have a sum b. Use mental math to determine a good place to start.

Algebra 1Algebra 1


Lesson 9-8

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 Find the product ac.

Step 3: Factors Sum 4 • 18 4 + 18 = 22 6 • 12 6 + 12 = 18 8 • 9 8 + 9 = 17

Find two factors of ac that have sum b. Use mental math to determine a good place to start.

Algebra 1Algebra 1


Lesson 9-8

(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).

Algebra 1 Lesson 9-1 Adding and Subtracting Polynomials.

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