THE FACTOR THEOREM AND ITS CONVERSEReynaldo B. Pantino, T2

Objectives:• To identify whether a

given factor is a factor of a polynomial function.

• To determine the factor of a polynomial function.

• To find f(x) when the roots or the zeros are given.

Just for a moment!Is 3 is a factor of 27? If yes, what makes it a factor of 27?

Just for a moment!Consider this pie graph below. Is each part of the figure is equally divided? What does it says?

Questions to answers:Is factor a divisor?

Is quotient a factor?

If P(x) is divided by (x –c)

then what can you say

about the remainder so

that (x – c ) is a factor

of P(x)?

Let us discuss the following:The Remainder Theorem states that when

the polynomial P(x) is divided by x – c, the remainder is P(c).

Example: When P(x) = x3 – x2 – 4x + 4 is divided by x – 2, the remainder is

0.

That is, P(2) = 0.

by remainder theorem

Let us discuss the following:The Remainder Theorem states that when

the polynomial P(x) is divided by x – c, the remainder is P(c).

Example: When P(x) = x3 – x2 – 4x + 4 is divided by x – 2, we have;by synthetic

division

22 11 -1-1 -4-4 44

11 11 -2-2 0022 22 -4-4

remainder

Let us discuss the following:

Notice that P(c) = 0, using synthetic division

P(x) = (x – c) ● Q(x) + R becomes

P(x) = (x – c) ● Q(x) + 0

P(x) = (x – c) ● Q(x).

22 11 -1-1 -4-4 44

11 11 -2-2 0022 22 -4-4

remainder

Remember that:FACTOR THEOREM

Let P(x) be a polynomial. If P(c) = 0, where c is a real number, then (x – c) is a factor of P(x). Conversely, if (x – c) is a factor of P(x), then P(c) = 0.

Since the theorem has a converse, the proof consists of two parts.

a.) If (x – c) is a factor of P(x), then P(c) = 0.

b.) If P(c) = 0, then (x – c) is a factor of P(x).

Remember that:Proof:

(a)

Suppose (x – c) is a factor of P(x), then P(x) = (x – c) ● Q(x). Since the equation is an identity and is true for any value of x, then it must be true for x = c. Then;

P(c) = (c – c) ● Q(x)

P(c) = 0 ● Q(x)

P(c) = 0

Remember that:Proof:

(b)

Suppose P(c) = 0. By remainder theorem, when P(x) is divided by (x – c), the remainder (R) = P(c) = 0. Then;

P(x) = (x – c) ● Q(x) + 0

P(x) = (x – c) ● Q(x)

Therefore, (x – c) is a factor of P(x).

Illustrative Examples:1. Show that x + 1 is a factor of 2x3 + 5x2 – 3.

Solution:

Let P(x) = 2x3 + 5x2 – 3

P(-1) = 2(-1)3 + 5(-1)2 – 3

P(-1) = -2 + 5 – 3

P(-1) = 0

By Factor theorem, x + 1 is a factor of

2x3 + 5x2 – 3.

Illustrative Examples:2. Show that x - 2 is a factor of

x4 + x3 – x2 – x - 18.

Solution:

Let P(x) = x4 + x3 – x2 – x - 18

P(2) = (2)4 + (2)3 – (2)2 – (2) – 18

P(2) = 16 + 8 – 4 – 2 – 18

P(2) = 0

By Factor theorem, x – 2 is a factor of

x4 + x3 – x2 – x - 18.

Illustrative Examples:3. Find a polynomial function of minimum degree whose

zeros are –2, 1, –1.

Solution:

By factor theorem, the polynomial must have the following as factors, (x + 2) (x – 1) and (x + 1)

Thus; P(x) = (x + 2)(x – 1)(x + 1)

= (x + 2)(x2 – 1)

= x3 + 2x2 – x – 2

Test Yourself: A - Use the factor theorem to determine whether the first polynomial is a factor of the second.

1. (x + 1); x3 + x2 + x + 1

2. (x + 2); x8 + 2x7 + x + 2

3. (a – 1); a3 – 2a2 + a – 2

4. (x – 2); 4x3 – 3x2 – 8x + 4

5. (y – 2); 3y4 – 6y3 – 5y + 10

Test Yourself: B – Find a polynomial function with integral coefficients that has the given numbers as roots.

1.) 0, 1, - 2

2.) 2, -1 , -2

3.) 1, 1, 3

4.) 1/2 , 1, -1, 2

5.) 0, 1/2, -1, 2

Exercises:1.) Find the value of k so that polynomial x – 2 is the factor of 2x3 – kx - 3.

2.) A. Tell whether the second polynomial is a factor of the first .

a. P(x) = 3x3 – 8x2 + 3x + 2; (x – 2)

b. P(x) = 2x4 + x3 + 2x + 1; (x + 1)

c. P(x) = x3 + 4x2 + x – 6; (x + 3)

d. G(x) = 4x3 – 6x2 + 2x + 1; (2x – 1)

e. H(x) = x3 – 6x2 + 3x + 10; (x – 1)

>Let’s play<Determine the value of k which is necessary to meet the given condition.

(x – 2) is a factor of 3x3 – x2 – 11x + k

(x + 3) is a factor of 2x5 + 5x4 + 3x3 + kx2 – 14x + 3

(x + 1) is a factor of –x4 + kx3 – x2 + kx + 10

(x + 2) is a factor of x3 + x2 + 5x + k

(x – 1) is a factor of x3 – x2 – 4x + k

(x – 5) is a factor of x3 – 3x2 – kx - 5

(x + 1) is a factor of 3x3 + kx2 – x – 2

(x + 4) is a factor of kx3 + 4x2 – x - 4

( x + 5) is a factor of kx2 + 4x - 5

(x – 2) is factor of x3 + 3x2 – kx + 2