# Mathematics - Class 9 / Grade 9

- Polynomials - Factor theorem
### 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 9^{th}Grade, Factor Theorem Practice PageFactor 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 x^{3}+ 4x^{2}+ x – 6.**Solution:**Let p(x) = x^{3}+ 4x^{2}+ 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) = x

^{3}+ 4x^{2}+ x – 6P(1) = 1

^{3}+ 4 x 1^{2}+ 1 – 61 + 4 + 1 – 6 = 0

Hence, (x-1) is a factor of p(x) = x

^{3}+ 4x^{2}+ x – 6**Example 2:**Find the remainder when 2x^{3}- 9x^{2}+ x+12 when divided by x - 1.**Solution:**Let p(x) = 2x^{3}- 9x^{2}+ x+12 be the given polynomial.Now,

P(x) = 2x

^{3}- 9x^{2}+ x+12P(1) = 2x1

^{3}- 9 x (1)^{2}+ 1 + 122 - 9 + 1 + 12 = 15 – 9 = 6

Therefore, 6 is the remainder when 2x

^{3}- 9x^{2}+ x+12 is divided by x - 1.

**Example 3:**Without actual division prove that 4x^{4}– 2x^{3}- 6x^{2}+ x + 1 is exactly divisible by 2x^{2}+ x - 1.**Solution:**Let f(x) = 4x^{4}– 2x^{3}- 6x^{2}+ x + 1 and g(x) = 2x^{2}+ x - 1 be the given polynomials.We have, g(x) = 2x

^{2}+ x - 1 = 2x^{2}+ 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) = 4x

^{4}– 2x^{3}- 6x^{2}+ x + 1f(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 x^{3}+ ax^{2}- 2x + a + 4.**Solution:**Let p(x) = x^{3}+ ax^{2}- 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-a

^{3}+ a^{3}+ 2a + a + 4 = 03a + 4 = 0

a = -4/3

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

**Example 5:**If ax^{3}+ bx^{2}+ 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) = ax

^{3}+ bx^{2}+ 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 = 48a + 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 x^{2}– 1 is a factor of ax^{4}+ bx^{3}+ cx^{2}+dx + e, show that a + c + e = b + d = 0**Solution:**Let p(x) = ax^{4}+ bx^{3}+ cx^{2}+dx + e be the given polynomial. Then,x

^{2}– 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 x^{3}- 3x^{2}– 12x + 19 so that the result is exactly divisible by x^{2}+ x – 6.**Solution:**Let p(x) = x^{3}- 3x^{2}– 12x + 19 and q(x) = x^{2}+ 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) = x

^{3}- 3x^{2}– 12x + 19 + ax + bWe have, q(x) = x

^{2}+ x – 6= x^{2}+ 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

x

^{3}- 3x^{2}– 12x + 19 + ax + b = 0(2)

^{3}– 3(2)^{2}– 12(2) + 19 + a(2) + b = 08 – 12 – 24 + 19 + 2a + b = 0

2a + b = 9 (i)

and x

^{3}- 3x^{2}– 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, x

^{3}- 3x^{2}– 12x + 19 will be divisible by x^{2}+ x – 6, if 2x + 5 is added to it.**Download to practice offline.**