7.3 Products and Factors of Polynomials

Objectives: Multiply polynomials, and divide one polynomial by another by using long

division and synthetic division.

Standard: 2.8.11.S. Analyze properties and relationships of polynomials.

2, 1, 3x x x

( 2)( 1)( 3)x x x

2

3 2 2

3 2

( 2)( 3)

2 3 3 6

2 5 6

x x x

x x x x x

x x x

Example 1. Making an open-top box out of a single rectangular sheet involves cutting and folding square flaps at each of the corners. These flaps are then pasted to the adjacent side to

provide reinforcement for the corners.The dimensions of the rectangular sheet and the square flaps

determine the volume of the resulting box. For the 12-inch-by-16-inch sheet shown, the volume function is V(x) = x(16 – 2x)(12 –

2x), where x is the side length in inches of the square flap.Write the volume function for the open-top box as a

polynomial function in standard form.

V(x) = x(16 – 2x)(12 – 2x)

(16x – 2x2) (12 – 2x)

192x – 32x2 – 24x2 + 4x3

4x3 – 56x2 + 192x

Ex 2A. Write f(x) = 2x2 (x2 + 2) (x - 3) as a polynomial in standard form.

(2x4 + 4x2) (x – 3)

2x5 – 6x4 + 4x3 – 12x2

Ex. 2B Write as a polynomial in standard form.

Just as a quadratic function is factored by writing it as the product of 2 factors, a Polynomial Expression is in FACTORED

FORM when it is written as the product of 2 or more factors.

Ex 3. Factor each polynomial.• x3 - 5x2 - 6x

• x3 + 4x2 +2x + 8

• x3 – 9x x (x2 – 9) x (x + 3) (x – 3)

• x3 – x2 + 2x – 2 x2 (x – 1) + 2 (x – 1) (x2 + 2) ( x – 1)

c. x3 + 1000

d. x3 – 125

(x + 10) (x2 – 10x + 100)

(x – 5) (x2 + 5x + 25)

Factor Theoremx – r is a factor of the polynomial expression

that defines the function P if and only if r is a solution of P(x) = 0, that is, if

and only if P(r) = 0.With the Factor Theorem, you can test for linear

factors involving integers by using substitution.

Use substitution to determine whether x + 3 is a factor of x3 – 3x2 – 6x + 8.

• Solution is x = - 3.

• (-3)3 – 3(-3)2 – 6(-3) + 8• -27 – 27 + 18 + 8

• -28• No, its not a factor.

DIVIDING POLYNOMIALS BY SYNTHETIC OR LONG DIVISIONA polynomial can be divided by a divisor of the form x – r (FIRST POWER)

by using long division or a shortened form of long division called synthetic division.

You can only use Synthetic Division if you’re dividing by a LINEAR factor (x+ a)

* Find the quotient.(x3 + 3x2 – 13x – 15) ÷ (x2 – 2x – 3)

The quotient is: x + 5

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

synthetic division to factor x3 - 13x – 12.

The Quotient is x2 - 3x - 4x2 - 3x – 4

(x - 4)(x + 1)(x + 3)

(zeros are x = 4, x = -1, and x = -3)

Remainder TheoremIf the polynomial expression that

defines the function of P is divided by x – a, then the

remainder

Ex 11. Given P(x) = 3x3 + 2x2 – 3x + 4, find P(3).

P(3) = 94

Ex 12. Given P(x) = 3x3 - 4x2 + 9x + 5, find P(6) by using both synthetic division and substitution.

P(6) = 563