# FACTOR THEOREM FOR POLYNOMIALS

Factor Theorem for Polynomials :

In this section, you will learn how factor theorem can be used for polynomials.

Factor Theorem :

If p(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number then

(i) p(a)  =  0 implies (x - a) is a factor of p(x).

(ii) (x - a) is a factor of p(x) implies p(a)  =  0.

Note :

(i) (x-a) is a factor of p(x), if p(a)  =  0.

(ii) (x+a) is a factor of p(x), if p(-a)  =  0.

(iii) (ax+b) is a factor of p(x), if p(-b/a)  =  0.

(iv) (x-a)(x-b) is a factor of p(x), if p(a) = 0 and p(b) = 0.

## Factor Theorem for Polynomials - Practice Questions

Question 1 :

Determine the value of m, if (x + 3) is a factor of

x3 - 3x2 - mx + 24

Solution :

Let

f(x)  =  x3 - 3x2 - mx + 24

Equate the factor (x + 3) to zero.

x + 3  =  0

Solve for x.

x  =  -3

By factor theorem,

(x + 3) is factor of f(x), if f(-3)  =  0

Then,

f(-3)  =  0

(-3)3 - 3(-3)2 - m(-3) + 24  =  0

-27 - 3(9) + 3m + 24  =  0

-27 - 27 + 3m + 24  =  0

3m - 30  =  0

3m  =  30

m  =  10

Question 2 :

If both (x - 2) and (x - 1/2) are the factors of

ax2 + 5x + b,

then show that a  =  b.

Solution :

Let p(x)  =  ax2 + 5x + b

Equate the factor (x - 2) to zero.

x - 2  =  0

Solve for x.

x  =  2

By factor theorem,

(x - 2) is factor of p(x), if p(2)  =  0

Then,

p(2)  =  0

a(2)2 + 5(2) + b  =  0

4a + 10 + b  =  0

b  =  -4a - 10 -----(1)

Equate the factor (x - 1/2) to zero.

x - 1/2  =  0

Solve for x.

x  =  1/2

By factor theorem,

(x - 1/2) is factor of p(x), if p(1/2)  =  0

p(1/2)  =  0

a(1/2)2 + 5(1/2) + b  =  0

a/4 + 5/2 + b  =  0

a/4 + 10/4 + 4b/4  =  0

(a + 10 + 4b) / 4  =  0

a + 10 + 4b  =  0

From (1), substitute (-4a - 10) for b.

a + 10 + 4(-4a - 10)  =  0

a + 10 - 16a - 40  =  0

-15a - 30  =  0

-15a  =  30

a  =  -2 -----(2)

Substitute -2 for a in (1).

(1)-----> b  =  -4(-2) - 10

b  =  8 - 10

b  =  -2 -----(3)

From (2) and (3), we have

a  =  b

Question 3 :

If (x - 1) divides the polynomial

kx3 - 2x2 + 25x  - 26

without remainder, then find the value of k .

Solution :

Let

p(x)  =  kx3 - 2x2 + 25x  - 26

Because (x - 1) divides p(x) without remainder, (x - 1) is a factor of p(x).

Equate the factor (x - 1) to zero.

x - 1  =  0

Solve for x.

x  =  1

By factor theorem,

(x - 1) is factor of p(x), if p(1)  =  0

Then,

p(1)  =  0

k(1)3 - 2(1)2 + 25(1)  - 26  =  0

k - 2 + 25 - 26  =  0

k - 28 + 25  =  0

k - 3  =  0

k  =  3

Question 4 :

Check if (x + 2) and (x - 4) are the sides of a rectangle whose area is x2 - 2x - 8 by using factor theorem.

Solution :

If (x + 2) and (x - 4) are the sides of a rectangle, then the area of the rectangle is

(x + 2)(x - 4)

But, it is given that the area of the rectangle is

x2 - 2x - 8

Then,

x2 - 2x - 8  =  (x + 2)(x - 4)

From, the above equation, (x + 2) and (x - 4) must be the factors of x2 - 2x - 8.

Let's verify.

Let p(x)  =  x2 - 2x - 8

By factor theorem,

(x + 2) is factor of p(x), if p(-2)  =  0

(x - 4) is factor of p(x), if p(-2)  =  0

Then,

 p(-2)  =  (-2)2 - 2(-2) - 8  p(-2)  =  4 + 4 - 8 p(-2)  =  0 p(4)  =  42 - 2(4) - 8p(4)  =  16 - 8 - 8 p(4)  =  0

So, (x + 2) and (x - 4) are the factors of (x2 - 2x - 8).

Hence, (x + 2) and (x - 4) are the sides of the rectangle. After having gone through the stuff given above, we hope that the students would have understood how factor theorem can be used for polynomials.

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