## Topic outline

• ### Polynomials - Factor theorem

Tags: Factor Theorem Examples and Solutions for Grade 9, Factor Theorem Worksheet PDF for Class IX, Factor Theorem Questions and Answers for ninth class, Revision Notes for Factor Theorem for 9th Grade, Factor Theorem Practice Page

• Factor Theorem

We know that if q(x) divides p(x) completely, that means p(x) is divisible by q(x) or, q(x) is a factor of p(x).

Now we will study a theorem which will help us to determine whether a polynomial q(x) is a factor of a polynomial p(x) or not without doing the actual division. This theorem is known as the factor theorem.

Factor Theorem: If p(x) is a polynomial of degree greater than 1 i.e. n > 1 and a is any real number such that p(a) = 0, then (x - a) is a factor of p(x) or we can say if (x - a) is a factor of p(x), then p(a) = 0

This implies,

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

(ax – b) is a factor of a polynomial p(x) if p(b/a) = 0.

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

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

Example 1: Show that (x-1) is a factor of the polynomial x3 + 4x2 + x – 6.

Solution: Let p(x) = x3 + 4x2 + x – 6 be the given polynomial.

By factor theorem, (x-a) is a factor of a polynomial p(x) if p(a) = 0. Therefore, in order to prove that x – 1 is a factor of p(x), we have to show that p(1) = 0

Now,

P(x) = x3 + 4x2 + x – 6

P(1) = 13 + 4 x 12 + 1 – 6

1 + 4 + 1 – 6 = 0

Hence, (x-1) is a factor of p(x) = x3 + 4x2 + x – 6

Example 2: Find the remainder when 2x3- 9x2+ x+12 when divided by x - 1.

Solution: Let p(x) = 2x3- 9x2+ x+12 be the given polynomial.

Now,

P(x) = 2x3- 9x2+ x+12

P(1) = 2x13 - 9 x (1)2 + 1 + 12

2 - 9 + 1 + 12 = 15 – 9 = 6

Therefore, 6 is the remainder when 2x3- 9x2+ x+12 is divided by x - 1.

Example 3: Without actual division prove that 4x4 – 2x3 - 6x2 + x + 1 is exactly divisible by 2x2 + x - 1.

Solution: Let f(x) = 4x4 – 2x3 - 6x2 + x + 1 and g(x) = 2x2 + x - 1 be the given polynomials.

We have, g(x) = 2x2 + x - 1 = 2x2 + 2x -1x - 1 = 2x(x+1)-1(x+1) = (2x-1) (x+1)

(2x-1) and (x+1) are factors of g(x)

In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that 2x-1 and x+1 are factors of f(x). For this it is sufficient to prove that f(1/2) = 0 and f(-1) = 0.

Now, f(x) = 4x4 – 2x3 - 6x2 + x + 1

f(1/2) = 4 x (1/2)4 – 2 x (1/2)3 - 6 x (1/2)2 + 1/2 + 1

= 1/4 – 1/4 – 3/2 + 1/2 + 1 = 0

f(-1) = 4 x (-1)4 – 2 x (-1)3 - 6 x (-1)2 - 1 + 1

= 4 + 2 - 6 – 1 + 1 = 0

(2x-1) and (x+1) are factors of f(x)

g(x) = (2x-1)(x+1) is a factors of f(x)

Hence, f(x) is exactly divisible by g(x)

Example 4: Find the value of a, if (x + a) is a factor of x3 + ax2 - 2x + a + 4.

Solution: Let p(x) = x3 + ax2 - 2x + a + 4 be the given polynomial. By factor theorem, (x + a) is a factor of p(x) if p(-a) = 0.

Now, p(-a) = 0

(-a)3 + a(-a)2 – 2(-a) + a + 4 = 0

-a3 + a3 + 2a + a + 4 = 0

3a + 4 = 0

a = -4/3

Hence, (x + a) is a factor of the given polynomial, if a = -4/3

Example 5: If ax3 + bx2 + x - 6 has x + 2 as a factor and leaves a remainder 4 when divided by (x-2), find the values of ‘a’ and ‘b’.

Solution: Let p(x) = ax3 + bx2 + x -6 be the given polynomial.

Now, (x+2) is a factor of p(x)

P(-2) = 0 (as x+2 = 0, x = -2)

a(-2)3 +b(-2)2 + (-2) – 6 = 0

-8a + 4b – 2 – 6= 0

-8a + 4b = 8

-2a + b = 2

It is given that p(x) leaves the remainder 4 when it is divided by (x-2).

Therefore, p(2) = 4 (as x – 2 = 0, x= -2)

a(2)3 + b(2)2 + 2 – 6 = 4

8a + 4b – 4 = 4

8a + 4b = 8

2a + b = 2

Adding (i) and (ii), we get

2b = 4

b = 2

Putting b = 2 in (i), we get

-2a + 2 = 2

-2a = 0

a = 0

Hence, a = 0 and b = 2

Example 6: If x2 – 1 is a factor of ax4 + bx3 + cx2 +dx + e, show that a + c + e = b + d = 0

Solution: Let p(x) = ax4 + bx3 + cx2 +dx + e be the given polynomial. Then,

x2 – 1 is a factor of p(x)

(x-1)(x+1) is a factor of p(x)

(x-1) and (x+1) are factors of p(x)

P(1) = 0 and p(-1) = 0 (as x – 1 = 0, x = 1 and x + 1 = 0, x = -1)

a + b + c + d + e = 0 and a – b + c – d + e = 0

Adding and subtracting these two equations, we get

2(a + c + e) = 0 and 2 (b + d) = 0

a + c + e = 0 and b + d = 0

a + c + e = b + d = 0

Example 7: What must be added to x3 - 3x2 – 12x + 19 so that the result is exactly divisible by x2 + x – 6.

Solution: Let p(x) = x3 - 3x2 – 12x + 19 and q(x) = x2 + x – 6. When p(x) is divided by q(x), the remainder is a linear expression in x. So, let r(x) = ax + b is added to p(x) so that p(x) + r(x) is divisible by q(x).

Let f(x) = p(x) + r(x). Then,

f(x) = x3 - 3x2 – 12x + 19 + ax + b

We have, q(x) = x2 + x – 6= x2 + 3x – 2x – 6 = x(x + 3) – 2(x + 3) = (x – 2) (x + 3)

Clearly, q(x) is divisible by (x – 2) and (x + 3) i.e. x – 2 and x + 3 are factors of q(x).

Therefore, f(x) will be divisible by q(x), if x – 2 and x + 3 are factors of f(x)

i.e., f(2) = 0 and f(-3) = 0

x3 - 3x2 – 12x + 19 + ax + b = 0

(2)3 – 3(2)2 – 12(2) + 19 + a(2) + b = 0

8 – 12 – 24 + 19 + 2a + b = 0

2a + b = 9 (i)

and x3 - 3x2 – 12x + 19 + ax + b = 0

(-3)3 – 3(-3)2 – 12(-3) + 19 + a(-3) + b = 0

-27 – 27 + 36 + 19 -3a + b = 0

1 – 3a + b = 0

-3a + b = -1 (ii)

Subtracting (i) from (ii) we get a = 2

Now substitute value of a = 2 in (i)

2 x 2 + b = 9

b = 5

Therefore, r(x) = ax + b

r(x) = 2x + 5

Hence, x3 - 3x2 – 12x + 19 will be divisible by x2 + x – 6, if 2x + 5 is added to it.