holt mcdougal algebra 2 6-3 dividing polynomials 6-3 dividing polynomials holt algebra 2 warm up...

Holt McDougal Algebra 2

6-3 Dividing Polynomials6-3 Dividing Polynomials

Holt Algebra 2

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Holt McDougal Algebra 2

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Warm UpDivide using long division.

1. 161 ÷ 7

2. 12.18 ÷ 2.1

3.

4.

2x + 5y

23

5.8

7a – b

Divide.

6x – 15y3

7a2 – aba

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Use long division and synthetic division to divide polynomials.

Objective

Holt McDougal Algebra 2

6-3 Dividing Polynomials

synthetic division

Vocabulary

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers.

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Divide using long division.

Example 1: Using Long Division to Divide a Polynomial

(–y2 + 2y3 + 25) ÷ (y – 3)

2y3 – y2 + 0y + 25

Step 1 Write the dividend in standard form, includingterms with a coefficient of 0.

Step 2 Write division in the same way you would when dividing numbers.

y – 3 2y3 – y2 + 0y + 25

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Notice that y times 2y2 is 2y3. Write 2y2 above 2y3.

Step 3 Divide.

2y2

–(2y3 – 6y2) Multiply y – 3 by 2y2. Then subtract. Bring down the next term. Divide 5y2 by y.

5y2 + 0y

+ 5y

–(5y2 – 15y) Multiply y – 3 by 5y. Then subtract. Bring down the next term. Divide 15y by y.

15y + 25

–(15y – 45)

70 Find the remainder.

+ 15

Multiply y – 3 by 15. Then subtract.

Example 1 Continued

y – 3 2y3 – y2 + 0y + 25

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Step 4 Write the final answer.

Example 1 Continued

–y2 + 2y3 + 25y – 3 = 2y2 + 5y + 15 +

70y – 3

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 1a

Divide using long division. (15x2 + 8x – 12) ÷ (3x + 1)

15x2 + 8x – 12

Step 1 Write the dividend in standard form, includingterms with a coefficient of 0.

Step 2 Write division in the same way you would when dividing numbers.

3x + 1 15x2 + 8x – 12

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 1a Continued

Notice that 3x times 5x is 15x2. Write 5x above 15x2.

Step 3 Divide.

5x

–(15x2 + 5x) Multiply 3x + 1 by 5x. Then subtract. Bring down the next term. Divide 3x by 3x.

3x – 12

+ 1

–(3x + 1)

–13Find the remainder.

Multiply 3x + 1 by 1. Then subtract.

3x + 1 15x2 + 8x – 12

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 1a Continued

Step 4 Write the final answer.

15x2 + 8x – 123x + 1 = 5x + 1 –

133x + 1

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 1b

Divide using long division. (x2 + 5x – 28) ÷ (x – 3)

x2 + 5x – 28

Step 1 Write the dividend in standard form, includingterms with a coefficient of 0.

Step 2 Write division in the same way you would when dividing numbers.

x – 3 x2 + 5x – 28

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Notice that x times x is x2. Write x above x2.

Step 3 Divide.

x

–(x2 – 3x) Multiply x – 3 by x. Then subtract. Bring down the next term. Divide 8x by x.

8x – 28

+ 8

–(8x – 24)

–4Find the remainder.

Multiply x – 3 by 8. Then subtract.

Check It Out! Example 1b Continued

x – 3 x2 + 5x – 28

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 1b Continued

Step 4 Write the final answer.

x2 + 5x – 28x – 3 = x + 8 –

4x – 3

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a).

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Divide using synthetic division.

Example 2A: Using Synthetic Division to Divide by a Linear Binomial

(3x2 + 9x – 2) ÷ (x – )

Step 1 Find a. Then write the coefficients and a in the synthetic division format.

Write the coefficients of 3x2 + 9x – 2.

13

For (x – ), a = .13

13

13

a =

13

3 9 –2

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Example 2A Continued

Step 2 Bring down the first coefficient. Then multiply and add for each column.

Draw a box around the remainder, 1 .13

13

3 9 –2

1

3

Step 3 Write the quotient.

3x + 10 +1 1

313

x –

10 131

133

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Example 2A Continued

3x + 10 +1 1

313

x – Check Multiply (x – ) 1

3

= 3x2 + 9x – 2

(x – ) 13

(x – ) 13

(x – ) 133x + 10 +

1 1313

x –

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Divide using synthetic division.

(3x4 – x3 + 5x – 1) ÷ (x + 2)

Step 1 Find a.

Use 0 for the coefficient of x2.

For (x + 2), a = –2.a = –2

Example 2B: Using Synthetic Division to Divide by a Linear Binomial

3 – 1 0 5 –1 –2

Step 2 Write the coefficients and a in the synthetic division format.

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Example 2B Continued

Draw a box around the remainder, 45.

3 –1 0 5 –1 –2

Step 3 Bring down the first coefficient. Then multiply and add for each column.

–6

3 45

Step 4 Write the quotient.

3x3 – 7x2 + 14x – 23 +45

x + 2Write the remainder over the divisor.

46–2814

–2314–7

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 2a

Divide using synthetic division.

(6x2 – 5x – 6) ÷ (x + 3)

Step 1 Find a.

Write the coefficients of 6x2 – 5x – 6.

For (x + 3), a = –3.a = –3

–3 6 –5 –6

Step 2 Write the coefficients and a in the synthetic division format.

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 2a Continued

Draw a box around the remainder, 63.

6 –5 –6 –3

Step 3 Bring down the first coefficient. Then multiply and add for each column.

–18

6 63

Step 4 Write the quotient.

6x – 23 +63

x + 3Write the remainder over the divisor.

–23

69

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 2b

Divide using synthetic division.

(x2 – 3x – 18) ÷ (x – 6)

Step 1 Find a.

Write the coefficients of x2 – 3x – 18.

For (x – 6), a = 6.a = 6

6 1 –3 –18

Step 2 Write the coefficients and a in the synthetic division format.

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 2b Continued

There is no remainder. 1 –3 –18 6

Step 3 Bring down the first coefficient. Then multiply and add for each column.

6

1 0

Step 4 Write the quotient.

x + 3

18

3

Holt McDougal Algebra 2

6-3 Dividing Polynomials

You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem.

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Example 3A: Using Synthetic Substitution

Use synthetic substitution to evaluate the polynomial for the given value.

P(x) = 2x3 + 5x2 – x + 7 for x = 2.

Write the coefficients of the dividend. Use a = 2.

2 5 –1 7 2

4

2 41

P(2) = 41

Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.P(2) = 2(2)3 + 5(2)2 – (2) + 7

P(2) = 41

3418

179

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Example 3B: Using Synthetic Substitution

Use synthetic substitution to evaluate the polynomial for the given value.

P(x) = 6x4 – 25x3 – 3x + 5 for x = – .

6 –25 0 –3 5

–2

6 7

13

–13

Write the coefficients of the dividend. Use 0 for the coefficient of x2. Use a = . 1

3P( ) = 71

3

2–39

–69–27

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 3a

Use synthetic substitution to evaluate the polynomial for the given value.

P(x) = x3 + 3x2 + 4 for x = –3.

Write the coefficients of the dividend. Use 0 for the coefficient of x2 Use a = –3.

1 3 0 4 –3

–3

1 4

P(–3) = 4

Check Substitute –3 for x in P(x) = x3 + 3x2 + 4.P(–3) = (–3)3 + 3(–3)2 + 4

P(–3) = 4

00

00

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Check It Out! Example 3b

Use synthetic substitution to evaluate the polynomial for the given value.

P(x) = 5x2 + 9x + 3 for x = .

5 9 3

1

5 5

151

5 Write the coefficients of the dividend. Use a = . 1

5

P( ) = 515

2

10

Holt McDougal Algebra 2

6-3 Dividing Polynomials

Example 4: Geometry ApplicationWrite an expression that represents the area of the top face of a rectangular prism when the height is x + 2 and the volume of the prism is x3 – x2 – 6x.

Substitute.

Use synthetic division.

The volume V is related to the area A and the height h by the equation V = A h. Rearrangingfor A gives A = . V

hx3 – x2 – 6x

x + 2A(x) =

1 –1 –6 0 –2

–2

1 0

The area of the face of the rectangular prism can be represented by A(x)= x2 – 3x.

06

0–3

Holt McDougal Algebra 2

6-3 Dividing PolynomialsCheck It Out! Example 4

Write an expression for the length of a rectangle with width y – 9 and area y2 – 14y + 45.

Substitute.

Use synthetic division.

The area A is related to the width w and the length l by the equation A = l w.

y2 – 14y + 45 y – 9

l(x) =

1 –14 45 9

9

1 0The length of the rectangle can be represented by l(x)= y – 5.

–45

–5

Holt McDougal Algebra 2

6-3 Dividing Polynomials

4. Find an expression for the height of a parallelogram whose area is represented by 2x3 – x2 – 20x + 3 and whose base is represented by (x + 3).

Lesson Quiz

2. Divide by using synthetic division. (x3 – 3x + 5) ÷ (x + 2)

1. Divide by using long division. (8x3 + 6x2 + 7) ÷ (x + 2)

2x2 – 7x + 1

194; –43. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1.

8x2 – 10x + 20 – 33

x + 2

x2 – 2x + 1 + 3

x + 2