# Factor theorem for polynomial | Factoring polynomials using factor theorem

If a polynomial f(x) is dividing by g(x) leaves remainder zero then g(x) is a factor of f(x). Here explained about factor theorem example with solution of factorise the polynomials by using factor theorem

## Factor Theorem Applications| Factor theorem example problems

Factor theorem state and proof purpose go through the below link

Statement and proof of Factor Theorem

Factor theorem practice problems

Example – 1 : Find the value of k, if (2x-3) is a factor of 2x3 – 9x2 + x + k

Solution: Let f(x) = 2x3 – 9x2 + x + k  &  g(x) = 2x-3

Since  g(x) is a factor of f(x)

This implies that f(3/2) = 0

Now put x = 3/2 in f(x)

-48/4 + K = 0

K = 12

Example – 2 :Find the value of “n” and ‘m” such that the polynomial x3 – nx2– 13x + m has (x  + 1) and ( x – 2) as factor.

Solution: Let f(x) =x3 – nx2– 13x + m

Since (x +1) is a factor of  f(x)

i.e f(-1) = 0

So f ( -1) =  – 1 -n +13 +m = 0

m -n = -12   – – – – – – – –  ( 1)

Since (x -2) is a factor of  f(x)

i.e f(2) = 0

So f ( 3) = 8 -4n  – 26 + m = 0

m -4n = 18   – – – – – – – – – (2)

Now from equations (1) and (2)

n = -10  & m = -22

Example – 3 : What much be added to x3 – 3x2 + 4x – 15 to obtain a polynomial which is exactly divisible by  x-3 ?

Solution: Let f (x) =  x3 – 3x2 + 4x – 15  &  g(x) = x -3

Since the degree of the g(x) is 1. So the degree of the remainder is 0

Let “b” be added to f(x), so that it may be exactly divisible by g(x)

i.e f(x) = (x3 – 3x2 + 4x -15) + b

Since f(x) is exactly divisible by (x-3)

So f(3) = 0

27 – 27 + 12 – 15 + b = 0

b = 3

i. e g(x) is a factor of f(x) while add constant of 3 to f(x)

Application of the factor theorem

Application 1

Prove that (x- a) is a factor of xn – an for any natural number of “n” by using factor theorem

Let f(x) = xn – a , g(x) = x –a Where n = Natural number

The remainder when f(x) is divide by g(x) is f(a)

Now put x = a in f(x) then

f(a) = an – a = 0

So f(a) = 0 for any natural number of “n”

i.e (x- a) is a factor of xn – an   where n N

Application 2

Prove that (x + a) is a factor of xn + an for any odd number of “n” by using factor theorem

Let f(x) = xn + a , g(x) = x + a Where n = odd number

The remainder when f(x) is divide by g(x) is f(-a)

Now put x = – a in f(x) then

f(-a) = (-a)n + a = – an + a = 0

So f(-a) = 0 for any  odd number of “n”

i.e (x + a) is a factor of xn + an   where n  = odd number

Example – 4 : Find the factors of the following polynomials

i) a3 +8           ii) a6 – b6             iii) 125a3 + 216b3

Solutions :

i )  a3 +8     = a3 + 23    where exponential is odd number

According to Application 2

(a +2 ) is factor of  a3 + 23

ii )  a6 – b6      where exponential is natural number

According to Application 1

(a – b ) is factor of  a6 – b6

iii )  125a3 + 216b3    = (5a)3 + (6b)3  where exponential is odd number

According to Application 2

(5a + 6b ) is factor of  125a3 + 216b3

Application 3 (Factorisation of a quatratic polynomial)

Factorise any quadratic polynomial ax2 + bx + c ( where a 0 and a,b & c are constants ), we have to write b as the sum of two numbers whose product is “ac”.

Proof:

Here a polynomial ax2 + bx + c ( where a 0 and a,b & c are constants )

Let its factors be   (rx + s) &  (px + q)  then

ax2 + bx + c = (rx + s) (px + q)  = rpx2 + (sp + rq)x + sq.

Comparing the coefficients of x2, we get that a = rp

Similarly comparing the coefficient of x, we get b = sp + rq

And on comparing the constant term, we get c = sq.

This shows that “b” is the sum of two numbers sp & rq, whose product is

(sp) (rp) =(rp)  (sq)  = a c

Therefore, to factorise  ax2 + bx + c , we have to write ‘ b ‘ as the sum of two numbers whose product is ‘ ac ‘

Example – 5 : Factorise using factor theorem 6x2 + 17x +5

Solution: By using the application -3

If we can find two numbers p and q such that p +q = 17 and pq =  6 x 5 = 30, we can get the factors

So, let us look for the factors of 30. Some are

1 & 30

3 & 10

2 & 15

Of these pairs 2 & 15 will give us  p +q = 2 + 15 = 17

So, 6x2 + 17x +5  = 6x2 + (15 +2)x +5

= 2x(3x +1) + 5(3x + 1)

= (2x +5) (3x +1)

Therefore (2x +5) & (3x +1) are factors of a polynomial  6x2 + 17x +5

Example – 6 : Factorise using factor theorem 2x2 + 5x + 3

Solution: By using the application -3

If we can find two numbers p and q such that p +q = 5 and pq =  2 x 3 = 6, we can get the factors

So, let us look for the factors of 6. Some are

1 & 6

2 & 3

Of these pairs 2 & 3 will give us  p +q = 2 +3 = 5

So, 2x2 + 5x + 3 = 2x2 + (2+3)x + 3

= x(2x +3) + (2x + 3)

= (2x +3) (x +1)

Therefore (2x +3) &  (x +1) are factors of a polynomial  2x2 + 5x + 3

Example – 7 :  Using factor theorem factorise x4 +4x3 -x2 – 16x -12

Solution: Let p(x) = x4 +4x3 -x2 – 16x -12

The constant term of p(x) is  -12 and its factors are

±1 , ±2, ±3, ±4, ±6 & ±12

So in beginning –  – –  – – –  till the value of p(x) os zero

Put x = 1 in p(x)

P(1) = 1 + 4 – 1 – 16 – 12 = 24 ≠ 0

i.e (x – 1) is not a factor for p(x)

Now put x = -1 in p(x)

P(-1) = 1 -4 -1 +16 – 12 = 17 – 17 = 0

i.e (x + 1) is a factor for p(x)

By dividing p(x) with (x+1),

We get the quotient x3 + 3 x2 – 4x – 12

So p (x) = (x +1) (x3 + 3 x2 – 4x – 12)

Let q(x) = x3 + 3 x2 – 4x – 12

Again we should starts to find q(2), q(-2) –  –  –  –  –

q(2) = 8 + 12 – 8 12 = 0

Therefore ( x- 2) is a factor of q(x)

By dividing q(x) with (x -2), we get quotient x2 + 5x + 6

q(x) = (x -2) (x2 + 5x + 6)

Finally

P(x) = x4 +4x3 -x2 – 16x -12

=  (x +1) (x3 + 3 x2 – 4x – 12)

= (x +1) (x -2) (x2 + 5x + 6)

= (x +1) (x -2) (x2 + 3 x +2x + 6)

= (x +1) (x -2) [ x(x + 3) + 2(x + 3)]

= (x +1) (x +2) (x -2)  (x + 3) Related Articles

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