Bell-ringer Geometry Text

7.3.2 Products and Factors of Polynomials

Objectives:

1).Divide one polynomial by another using long and synthetic division

2).Use the Remainder Theorem to solve problems

Definitons Divisor-expression that you divide by Dividend – the expression being divided Quotient - the answer after division is

complete Remainder - what is left over after dividing.

Refer to the example

Long Division with numbers

6478

Long Division With Numbers

31 238 2 87

- 21721

6

- 18626

8

- 24820

2 8

Find the quotient: (23828) ÷ (31)

3120

768 20/31 or 768 + 20/31

Practice with variables2x2 ÷ x 9y3÷ 3y

16w2 ÷4w 36x3÷2x

- (–14x + 56)

x – 4 x3 – 2x2 – 22x + 40

x2

- (x3 – 4x2)2x2 – 22x

+ 2x

- (2x2 – 8x) –14x + 40

– 14

– 16

x2 + 2x – 14 –x – 416

ExampleFind the quotient: (x3 – 2x2 – 22x + 40) ÷ (x – 4)

x – 4- 16

Example

x2 + x + 2 x3 –x2 + 0x - 4

x

- (x3 + x2 +2x)-2x2 – 2x - 4

- 2

- (-2x2 – 2x – 4) 0

Find the quotient: (x3 –x2 – 4) ÷ (x2 + x + 2)

Quotient: x – 2

Now You Find the quotient:

(x3 + 3x2 – 13x - 15) ÷ (x2 – 2x – 3)

Quotient: x + 5

Synthetic DivsionUse synthetic division to find the quotient:

(x3 + x2 – 9x - 9) ÷ (x -3)

13 1 -9 -9

Step 1: Write the coefficients of the polynomial, and the r-value of the divisor on the left

Example 2Use synthetic division to find the quotient:

(x3 + x2 – 9x - 9) ÷ (x -3)

13

1

Step 2: Draw a line and write the first coefficient under the line.

1 1 -9 -9

Example 2Use synthetic division to find the quotient:

(x3 + x2 – 9x - 9) ÷ (x - 3)

Step 3: Multiply the r-value, 3, by the number below the line and write the product below the next coefficient.

3

13

1 1 -9 -9

Example 2Use synthetic division to find the quotient:

(x3 + x2 – 9x - 9) ÷ (x - 3)

Step 4: Write the sum of 1 and 3 below the line.

3

134

1 1 -9 -9

Example 2Use synthetic division to find the quotient:

(x3 + x2 – 9x - 9) ÷ (x - 3)

Repeat steps 3 and 4.

3

134

123

1 1 -9 -9

Example 2Use synthetic division to find the quotient:

(x3 + x2 – 9x - 9) ÷ (x - 3)

Repeat steps 3 and 4.

3

134

123

90

1 1 -9 -9

Example 2

(x3 + x2 – 9x - 9) ÷ (x - 3)

3

134

123

90

The remainder is 0 and the resulting numbers are the coefficients of the quotient.

x2 + 4x + 3

Use synthetic division to find the quotient:

1 1 -9 -9

Synthetic Division v. long division

*Synthetic Division can only be used when dividing by a linear binomial of the form x –r.

**Otherwise long division must be used.

(x4 – 3x + 2x3 – 6) ÷ (x - 2)

2

124

88

1613

Use synthetic division to find the quotient:

1 2 0 -3

Example

-626

20

x3 + 4x2 + 8x + 13 + 20/(x-2)

Now You

(6x2 – 5x - 6) ÷ (x + 3)

-3

6-18-23

69

Use synthetic division to find the quotient:

6 -5 -6

63

6x -23 + 63/(x+3).

Example 3Given that 2 is a zero of P(x) = x3 – 3x2 + 4, use division to factor x3 – 3x2 + 4.

Since 2 is a zero, x = 2 , so x – 2 = 0 , which means x – 2is a factor of x3 – 3x2 + 4.

(x3 – 3x2 + 4) ÷ (x – 2)

Method 1 Method 2

- (–2x + 4)

x – 2 x3 – 3x2 + 0x + 4

x2

- (x3 – 2x2)-x2 + 0x

- x

- (-x2 + 2x) –2x + 4

– 2

0

1

2

-1

-2

-2

-4

0

2 1 -3 0 4

x3 – 3x2 + 4 = (x – 2)(x2 – x – 2)

Practice

Given that -3 is a zero of P(x) = x3 – 13x - 12, use division to factor x3 – 13x – 12.

x3 -13x -12 = (x + 3)(x2 -3x -4)

Remainder Theorem

If the polynomial expression that defines the function of P is divided by x – a, then the remainder is the number P(a).

Example 4

Given P(x) = 3x3 – 4x2 + 9x + 5 find P(6) by using both synthetic division and substitution.

3

18

14

84

93

558

563

6 3 -4 9 5

Method 1 Method 2

P(6) = 3(6)3 – 4(6)2 + 9(6) + 5

= 3(216) – 4(36) + 54 + 5

= 648 – 144 + 54 + 5

= 563

Practice

Given P(x) = 3x3 + 2x2 + 3x + 1 find P(-2) using synthetic division and substitution.

Homework

p.446 #72 – 96 by 3’s