## Factor Theorem Applications| Factor theorem example problems

Statement and proof of Factor Theorem

**Factor theorem practice problems**

**Example – 1 :** Find the value of k, if (2x-3) is a factor of 2x^{3} – 9x^{2} + x + k

**Solution:** Let f(x) = 2x^{3} – 9x^{2} + 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 x^{3} – nx^{2}– 13x + m has (x + 1) and ( x – 2) as factor.

**Solution**: Let f(x) =x^{3} – nx^{2}– 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 x^{3} – 3x^{2 }+ 4x – 15 to obtain a polynomial which is exactly divisible by x-3 ?

**Solution:** Let f (x) = x^{3} – 3x^{2 }+ 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) = (x^{3} – 3x^{2} + 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 x^{n} – a^{n} for any natural number of “n” by using factor theorem

Let f(x) = x^{n} – a^{n } , 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) = a^{n} – a^{n } = 0

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

i.e (x- a) is a factor of x^{n} – a^{n }where n N

**Application 2
**

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

Let f(x) = x^{n} + a^{n } , 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^{n } = – a^{n} + a^{n } = 0

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

i.e (x + a) is a factor of x^{n} + a^{n }where n = odd number

**Example – 4 :** Find the factors of the following polynomials

i) a^{3} +8 ii) a^{6} – b^{6} iii) 125a^{3} + 216b^{3}

**Solutions :**

i ) a^{3} +8 = a^{3} + 2^{3} where exponential is odd number

According to Application 2

(a +2 ) is factor of a^{3} + 2^{3}

ii ) a^{6} – b^{6} where exponential is natural number

According to Application 1

(a – b ) is factor of a^{6} – b^{6}

iii ) 125a^{3} + 216b^{3} = (5a)^{3} + (6b)^{3} where exponential is odd number

According to Application 2

(5a + 6b ) is factor of 125a^{3} + 216b^{3}

**Application 3 (Factorisation of a quatratic polynomial)
**

Factorise any quadratic polynomial **ax ^{2} + 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 **ax ^{2} + bx + c** ( where a 0 and a,b & c are constants )

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

ax^{2} + bx + c = (rx + s) (px + q) = rpx^{2} + (sp + rq)x + sq.

Comparing the coefficients of x^{2}, 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 ax^{2} + bx + c , we have to write ‘ b ‘ as the sum of two numbers whose product is ‘ ac ‘

**Example – 5 :** Factorise using factor theorem 6x^{2} + 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, 6x^{2 }+ 17x +5 = 6x^{2} + (15 +2)x +5

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

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

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

**Example – 6 : **Factorise using factor theorem 2x^{2} + 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, 2x^{2} + 5x + 3 = 2x^{2} + (2+3)x + 3

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

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

Therefore (2x +3) & (x +1) are factors of a polynomial 2x^{2} + 5x + 3

**Example – 7 :**Using factor theorem factorise x

^{4}+4x

^{3}-x

^{2}– 16x -12

**Solution:** Let p(x) = x^{4} +4x^{3} -x^{2} – 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 x^{3} + 3 x^{2} – 4x – 12

So p (x) = (x +1) (x^{3} + 3 x^{2} – 4x – 12)

Let q(x) = x^{3} + 3 x^{2} – 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 x^{2} + 5x + 6

q(x) = (x -2) (x^{2} + 5x + 6)

Finally

P(x) = x^{4} +4x^{3} -x^{2} – 16x -12

= (x +1) (x^{3} + 3 x^{2} – 4x – 12)

= (x +1) (x -2) (x^{2} + 5x + 6)

= (x +1) (x -2) (x^{2} + 3 x +2x + 6)

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

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

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