Obj. 5 Synthetic Division

Unit 2 Quadratic and Polynomial Functions

Concepts and Objectives

� Synthetic Division (Obj. #5)

� Review performing synthetic division

� Evaluate polynomial functions using the remainder

theorem

Quadratic Functions

� A polynomial function of degree n, where n is a

nonnegative integer, is a function defined by an

expression of the form

( ) −

−= + + + +1

1 1 0...n n

n nf x a x a x a x a

� A function f is a quadratic function if

where a, b, and c are real numbers, and a ≠ 0.

( ) = + +2f x ax bx c

Dividing Polynomials

� Let f(x) and g(x) be polynomials with g(x) of degree one

or more, but of lower degree than f(x). There exist

unique polynomials q(x) and r(x) such that

where either r(x) = 0 or the degree of r(x) is less than the

( ) ( ) ( ) ( )= +i f x g x q x r x

where either r(x) = 0 or the degree of r(x) is less than the

degree of g(x).

Dividing Polynomials

� For example, could be evaluated as− −

−

3 2

2

3 2 150

4

x x

x

− − + −2 3 2

3

4 3 2 0 150

x

x x x x

− + +3 23 0 12x x x

−2

or

− + +3 23 0 12x x x

− + −22 12 150x x+ −22 0 8x x

−12 158x

−− +

−2

12 1583 2

4

xx

x

Dividing Polynomials

� Using the division algorithm, this means that

( )( )− − = − − + −3 2 23 2 150 4 3 2 12 158x x x x x

( )f x ( )g x ( )q x ( )r x( ) ( ) ( ) ( )

Dividend = Divisor • Quotient + Remainder

Synthetic Division

� A shortcut method of performing long division with

certain polynomials, called synthetic division, is used

only when a polynomial is divided by a binomial of the

form x – k, where the coefficient of x is 1.

� To use synthetic division:� To use synthetic division:

The answers are the coefficients of the quotient.

−1 1 0 ...

n nk a a a a

−

−+ + + +

=−

1

1 1 0...n n

n na x a x a x a

x k

an

kan

−+1n na ka …

Synthetic Division

� Example: Use synthetic division to divide

− + −

−

3 24 15 11 10

3

x x x

x

− −3 4 15 11 10− −3 4 15 11 10

4

12

–3

–9

2

6

–4

−− + +

−

2 44 3 2

3x x

x

Synthetic Division

� Example: Use synthetic division to divide

+ + − +

+

4 3 25 4 3 9

3

x x x x

x

− −3 1 5 4 3 9− −3 1 5 4 3 9

1

–3

2

–6

–2

6

3

–9

0

+ − +3 22 2 3x x x

Remainder Theorem

� The remainder theorem:

Example: Let . Find f(–3).

If the polynomial f(x) is divided by x – k, then the

remainder is equal to f(k).

( ) = − + − −4 23 4 5f x x x x� Example: Let . Find f(–3).( ) = − + − −4 23 4 5f x x x x

− − − −3 1 0 3 4 5

–1

3

3

–9

–6

18

14

–42

–47

( )− = −3 47f

Potential Zeros

� A zero of a polynomial function f is a number k such that

f(k) = 0. The real number zeros are the x-intercepts of

the graph of the function.

� The remainder theorem gives us a quick way to decide if

a number k is a zero of a polynomial function defined by a number k is a zero of a polynomial function defined by

f(x). Use synthetic division to find f(k) ; if the remainder

is 0, then f(k) = 0 and k is a zero of f(x) .

Potential Zeros

� Example: Decide whether the given number k is a zero

of f(x):

( ) = − − + + = −4 3 24 14 36 45; 3f x x x x x k

− − −3 1 4 14 36 45

Since the remainder is zero, –3 is a zero of the function.

− − −3 1 4 14 36 45

1

–3

–7

21

7

–21

15

–45

0

Homework

� Page 326: 5-50 (×5), 54, 55

� HW: 10, 30, 40, 50, 54