Ch. 6.3 Dividing Polynomials

Divide x2 + 2x – 30 by x – 5. 

ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3

Dividing PolynomialsDividing Polynomials

– 30 Subtract: (x2 + 2x) – (x2 – 5x) = 7x. Bring down –30.

x Divide = x.x – 5 x2 + 2x – 30

x2

x

x2 – 5x Multiply: x(x – 5) = x2 – 5x 7x

Repeat the process of dividing, multiplying, and subtracting.

5 Subtract: (7x – 30) – (7x – 35) = 5.

The quotient is x + 7 with a remainder of 5, or simply x + 7, R 5.

7x – 35 Multiply: 7(x – 5) = 7x – 35.

x + 7 Divide = 7.x – 5 x2 + 2x – 30

x2 – 5x7x – 30

7xx

6-3

P. 315 Check Understanding 1

Determine whether x + 2 is a factor of each polynomial.

ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3

Dividing PolynomialsDividing Polynomials

a. x2 + 10x + 16 b. x3 + 7x2 – 5x – 6

Since the remainder is zero,x + 2 is a factor of x2 + 10x + 16.

x + 8x + 2 x2 + 10x + 16

x2 + 2x8x + 16 8x + 16

0

x2 + 5x – 15x + 2 x3 + 7x2 – 5x – 6

x3 + 2x2

5x2 – 5x5x2 + 10x

–15x – 6–15x – 30

24

Since the remainder = 0, x + 2 is not a factor of x3 + 7x2 – 5x – 6.

/

6-3

p. 315 Check Understanding # 2a & B

Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3.

ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3

Dividing PolynomialsDividing Polynomials

Step 1: Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form.

Step 2: Bring down the coefficient.

Write x – 3 5x3 – 6x2 + 4x – 1

as 3 5 –6 4 –1

Bring down the 5.3 5 –6 4 –1 This begins the quotient.

5

6-3

(continued)

ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3

Dividing PolynomialsDividing Polynomials

Multiply 3 by 5. Write3 5 –6 4 –1 the result under –6.x 15

5 9 Add –6 and 15.

Step 3:  Multiply the first coefficient by the new divisor. Write the result under the next coefficient. Add.

Step 4:  Repeat the steps of multiplying and adding until the remainder is found.

The quotient is 5x2 + 9x + 31, R 92.

3 5 –6 4 –115 27 93

5 9 31 92

5x2 + 9x + 31  Remainder

6-3

P. 316 Check understanding # 3

The volume in cubic feet of a shipping carton is V(x) = x3 – 6x2 + 3x + 10. The height is x – 5 feet.

ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3

Dividing PolynomialsDividing Polynomials

a. Find linear expressions for the other dimensions. Assume that the length is greater than the width.

b. If the width of the carton is 4 feet, what are the other two dimensions?

x2 – x – 2 = (x – 2)(x + 1) Factor the quotient.The length and the width are x + 1 and x – 2, respectively.

x – 2 = 4  Substitute 4 into the expression for width. Find x.x = 6Since the height equals x – 5 and the length equals x + 1, the height is 1 ft. and the length is 7 ft.

5 1 –6 3 10 Divide. 5 –5 –10

1 –1 –2 0

x2 – x – 2  Remainder

6-3

p. 317 Check understanding # 4A & B

Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9.

ALGEBRA 2 LESSON 6-3ALGEBRA 2 LESSON 6-3

Dividing PolynomialsDividing Polynomials

By the Remainder Theorem, P(3) equals the remainder when P(x) is divided by x – 3.

The remainder is 21, so P(3) = 21.

3 1 –2 0 1 –9 3 3 9 30

1 1 3 10 21

6-3

Ch. 6.3 p. 318 # 1, 2, 7 – 10, 13, 18, 19, 21, 23, 28, 29