1

2

A polynomial function is a function of the form

11 1 0( ) , 0n n

n n nf x a x a x a x a a

where n is a nonnegative integer and each ai (i = 0,1,…, n)

is a real number. The polynomial function has a leading coefficient an and degree n.

Def.:

3

+ 2 2 3 1 2 xxx

Ex1: Divide x2 + 3x – 2 by x + 1 and check the answer.

x

x2 + x2x – 22x + 2

– 4

remainder

Check:

xx

xxx

22 1.

xxxx 2)1(2.

xxxxx 2)()3( 22 3.

22

2 x

xxx4.

22)1(2 xx5.

4)22()22( xx6.

(x + 2)

quotient

(x + 1)

divisor

+ (– 4)

remainder

= x2 + 3x – 2

dividend

Answer: x + 2 +1x

– 4

Dividing Polynomials

4

Notes:

)(

)()(

)(

)(

xD

xRxQ

xD

xP

• P(x): Dividend

• D(x): Divisor

• Q(x): Quotient

• R(x): Remainder

1)

2) The degree of R(x) <the degree of D(x)

5

5210732

:Divide2

234

xxxxx

Ex2

Ex3(HW)

5230232

:2

234

xxxxx

Divide

6

16

Synthetic division is a shorter method of dividing polynomials.

This method can be used only when the divisor is of the form

x – c. It uses the coefficients of each term in the dividend.

Ex4: Divide 3x2 + 2x – 1 by x – 2 using synthetic division.

3 2 – 12

Since the divisor is x – 2, c = 2.

3

1. Bring down 3

2. (2 • 3) = 6

6

8 15

3. (2 + 6) = 8

4. (2 • 8) = 16

5. (–1 + 16) = 15coefficients of quotient remainder

value of c coefficients of the dividend

3x + 8Answer: 2

x15

7

Ex5:

Use synthetic division to divide 10112 24 xxx

by 3x

Ex6:

Use synthetic division to divide 4272 35 xxx

by 2x

8

•Remainder Theorem

If a polynomial P (x) is divided by x –c , then the remainder equals

P(c).

Ex7: Using the remainder theorem, evaluate P(x) = x 4 – 4x – 1

when x = 3.

9

1 0 0 – 4 – 13

1

3

3 9

6927

23 68

The remainder is 68 at x = 3, so P(3) = 68.

You can check this using substitution: P(3) = (3)4 – 4(3) – 1 = 68.

value of x

Note:

9

Ex8:

If P(x)=211x4-212x3 +212x2 +210x-3, then find P(1/211).

Q69/289:

Find the remainder of 5x48+6x10-5x+7 divided by x-1.

Ex9:(HW)

Find the remainder of P(x)=x103+x102+x101+x100 divided by x+i.

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Factor Theorem A polynomial P(x) has a factor (x – c) if and only if P(c) = 0.

Ex10: Show that (x + 2) and (x – 1) are factors of

P(x) = 2x 3 + x2 – 5x + 2.

6

2 1 – 5 2– 2

2

– 4

– 3 1

– 2

0

The remainders of 0 indicate that (x + 2) and (x – 1) are factors.

– 1

2 – 3 11

2

2

– 1 0

The complete factorization of P is (x + 2)(x – 1)(2x – 1).

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A real number c is a zero of P (x) if and only if P(c) = 0.

A polynomial function of degree n has at most n real zeros.

Real Zeros of Polynomial Functions

If P(x) is a polynomial and c is a real number then the following statements are equivalent.

1. x = c is a zero of P.

2. x = c is a solution of the polynomial equation P (x) =0.

3. (x – c) is a factor of the polynomial P (x).

4. (c, 0) is an x-intercept of the graph of P (x).

and we have P(x)=(x-c) Q(x), where Q(x) is a polynomial of degree< degree of P(x) by 1 called the reduced polynomial

12

Ex11: Verify that (x+4) is a factor of P(x) = 3x4+11x3-6x2-6x+8. and write P(x) as the product of (x+4) and the reduced polynomial Q(x).

Ex12:Prove that for any positive odd integer n, P(x) = xn +1 has x-1 as a factor.

Ex13: If x-i is a factor of the polynomial P(x) = 7x171-8x172-9x173+kx174 , then find the value of k .