# Factorisation Quadratic Polynomial

## Polynomial of Class 9

## FACTORISATION QUADRATIC POLYNOMIAL

For factorisation of a quadratic expression ax^{2} + bx + a where a ≠ 0, there are two method.

###
**(a) By Method of Completion of Square :**

In the form ax^{2} + bx + c where a ≠ 0, firstly we take ‘a’ common in the whole expression then factorise by converting the expression as the difference of two squares.

**Question **Factorize ^{x2}– 31x + 220.

**Solution: **x^{2}– 31a + 220

**Question **Factorize: –10x^{2} + 31x – 24

**Solution: **

**Question **81a^{2}b^{2}c^{2}+ 64a^{6}b^{2}– 144a^{4}b^{2}c

**Solution: **81a^{2}b^{2}bc^{2} + 64a^{6}b^{2}– 144a^{4}b^{2}c

= [9abc] ^{2}– 2 [9abc][8a^{3}b] + [8a^{3}b]^{2}

= [9abc –8a^{3}b]^{2} = a^{2}b^{2 }[9c – 8a^{2}]^{2} Ans.

**Question **

** **

###
**(b) By Splitting the Middle Term:**

In the quadratic expression ax^{2} + bx + c, where a is the coefficient of x^{2}, b is the coefficient of x and c is the constant term. In the quadratic expression of the form x^{2 }+ bx + c, a = 1 is the multiple of x^{2} and another terms are the same as above.

**There are four types of quadratic expression with respect to to the sign change of coefficients of x and constant term :**

(i) ax^{2} + bx + c (ii) ax^{2}– bx + c

(iii) ax^{2}– bx – c (iv) ax^{2} + bx – c

Splitting middle term of ax^{2} + bx +c.

In this case, we find integers s and t such that **s + t = b** and **st = ac.**

**Question **Factorize: .

**Solution: **Here

We try to split 5x into two parts whose sum is 5 and product is = –24

Clearly, 8 + (–3) = 5 and 8 × –3 = –24

Middle term 5x can be written as 8x - 3x

∴

**Question **Find the value of q, if ** X + 3** is a factor of

**Solution: ** Let be the given polynomial. According to question **(X+3) **is a factor of **p(x)**

∴ we have p(-3)= 0

⇒

Ex. Factorize: 2x^{2} + 12 + 35.

Sol. 2x^{2} + 12 + 35

Product ac = 70 & b = 12

.

**Question **Factorize: x^{2} – 14x + 24.

**Solution: **Product ac = 24 & b = -14

∴ Split the middle term as – 2 &– 2

⇒ x^{2}– 14x + 24 = x^{2}– 12 – 2x + 24

⇒ x(x – 12) – 2 (x – 12) = (x – 12)(x – 2) Ans.

**Question **Factorize:

**Solution: **

Product ac = – 48 & b = – 13 ∴ We split the middle term as – 16x + 3x.

=**1/24**[24x^{2}– 16x + 3x – 2]

=**1/24**[8x(3x – 2) + 1(3x – 2)]

=**1/24**(3x – 2)(8x + 1) Ans.

**Question ** Factorize:

**Solution: **

Ans.

**Question ** Factorize:

**Solution: ** The given expression is

We try to split 9 into two parts whose sum is 9 and product 18.

Clearly, 6 + 3 = 9 and 6 × 3 = 18.

∴

**Question ** Factorize: by splitting the middle term:

**Solution: ** In order to factorize we try to split **3√3** into two parts whose sum is **3√3** and product is 6.

So, we write the middle term **3√3** as **2√3x + √3x**

###
**(c) Factorisation by Using the Formula for the Difference of Two Squares:**

a^{2}– b^{2} = ( a + b) (a - b)

Ex. Factorise:4(2a + 3b – 4c) ^{2}– (a – 4b + 5c)^{2}

**Solution: **= 4(2a + 3b –4c)^{2}– (a – 4b + 5c) ^{2}

= [2(2a + 3b – 4c)]^{2}– (a – 4b + 5c)^{2}

= [4a + 6b – 8c + a – 4b + 5c] [4a + 6b – 8c – a + 4b – 5c]

= [5a + 2b – 3c] [3a + 10b – 13c] Ans.

**Question **Factorise :

**Solution: **

**Question **Factorize:

**Solution: **

Ans.

**Question **Factorize: x^{4} + x^{2}y^{2} + y^{4}.

Sol. x^{2} + x^{2}y^{2} + y^{4} = (x^{2}) 2 + 2.x^{2}.y^{2} + (y2)^{2}– x^{2}y^{2}

= x^{2} + y^{2})^{2}– (xy)^{2}

= (x^{2} + y^{2} + xy) (x^{2} + y^{2}– xy) Ans.

###
**(d) Factorization by Using Formula of a**^{3} + b^{3} and a^{3}– b^{3}:

^{3}+ b

^{3}and a

^{3}– b

^{3}:

Ex. Factorize: 64a^{13}b + 343ab^{13}.

**Solution: **64a^{13}b + 343ab^{13} = ab[64a^{12} + 343b^{12}]

= ab[(4a4)^{3 }+ (7b4)^{3}]

= ab[4a^{4} + 7b^{4}] [(4a4)2– (4a4) (7b4) + (7b4)2]

= ab[4a^{4} + 7b^{4}][16a^{8}– 28a^{4}b^{8} + 49b^{8}] Ans.

**Question ** Factorize: p^{3}q^{2}x^{4} + 3p^{2}qx^{3} + 3px^{2} + **X/q**– q^{2}r^{3}x

**Solution: **In above question, If we take common then it may become in the form of 3 + b^{3}.

p^{3}q^{2}x^{4} + 3p^{2}qx^{3} + 3px^{2} + **X/q**– q^{2}r^{3}x

= **X/q**[p^{3}q^{3}x^{3} + 3p^{2}q^{2}x^{2} + 3pqx + 1 – q^{3}r^{3}]

= **X/q**[(pqx)^{3} + 3(pqx)2 .1 + 3pqx . (1)^{2} + (1)^{3}– q^{3}r^{3}]

Let pqx = A & 1 = B

= **X/q** [A^{3} + 3A^{2}B + 3AB^{2} + B^{3}– q^{3}r^{3}]

= **X/q**[(pqx + 1)^{3}– (qr)^{3}] = **X/q**[pqx + 1 – qr][(pqx + 1)^{2} + (pqx + 1) qr + (qr)^{3}]

= **X/q**[pqx + 1 – qr][p^{2}q^{2}x^{2 }+ 1 + 2pqx + pq^{2}xr + qr + q^{2}r^{2}] Ans.

**Question **Factorize: x^{3}– 6x^{2} + 32

**Solution: **x^{3} + 32 – 6x^{2}

= x^{3} + 8 + 24 – 6x^{2}

= [(x)^{3} + (2)^{3}] + 6[4 – x^{2}]

= (x + 2)[x^{2}– 2x + 4] + 6[2 + x][2 – x]

= (x + 2) [x^{2}– 2x + 4 + 6(2 – x)]

= (x + 2)[x^{2}– 2x + 4 + 12 – 6x]

= (x + 2) [x^{2}– 8x + 16]

= (x + 2) (x – 4)^{2} Ans.

### (e) Integral Root Theorem :

If f(x) is a polynomial with integral coefficient and the leading coefficient is 1, then any integer root of f(x) is a factor of the constant term. Thus if f(x) = x^{3}– 6x^{2 }+ 11x - 6 has an Integral root, then it is one of the factors of 6 which are .

Now Infect f(1) = (1)^{3}– 6(1)2 + 11(1) – 6 = 1 – 6 + 11 – 6 =0

f(2) = (2)^{3}– 6(2)^{2} + 11(2) – 6 = 8 – 24 + 22 – 6 = 0

f(3) = (3)^{3}– 6(3)^{2} + 11(3) – 6 = 27 – 54 + 33 – 6 = 0

Therefore Integral roots of f(x) are 1,2,3.

### (f) Rational Root Theorem :

Let **b/c** be a rational fraction in lowest terms. If **b/c** is a root of the polynomial f(x) = with integral coefficients. Then b is a factor of constant term a0 and c is a factor of the leading coefficient an.

**For example :**If b/c is a rational root of the polynomial f(x) = 6x^{3} + 5x^{2}– 3x – 2, then the values of b are limited to the factors of –2 which are and the value of c are limited to the factors of 6, which are Hence, the possible rational roots of f(x) are Infect -1 is a Integral root and 2/3,-1/2 are the rational roots of f(x) = 6x^{3} + 5x^{2}– 3x – 2.

### NOTE:

(i) An nth degree polynomial can have at most n real roots.

(ii) Finding a zero or root of polynomial f(x) means solving the polynomial equation f(x) = 0. It follows from the above discussion that if f(x) = ax + b, a ≠ 0 is a linear polynomial, then it has only one root given by f(x) = 0 i.e. f(x) = ax + b = 0

⇒ ax = – b

⇒ x =

Thus a= is the only root of f(x) = ax + b.

**Question **If f(x) = 2x^{3}– 13x^{2} + 17x + 12 then find out the value of f(-2) & f(3).

**Solution: ** f(x) = 2x^{3}– 13x^{2 }+ 17x + 12

f(-2) = 2(–2)^{3}– 13(–2)^{2} + 17 (–2) + 12

= –16 – 52 – 34 + 12 = – 90 Ans.

f(3) = 2(3)^{3}– 13(3)^{2} + 17(3) + 12

= 54 – 117 + 51 + 12 = 0 Ans.

###
**(f) Factorisation of an Expression Reducible to a Quadratic Expression :**

**Question ** Factorize:- 8 + 9(a – b)^{6}– (a – b)^{12}

**Solution: **–8 + 9(a –b)^{6}– (a – b)^{12}

Let (a –b)6 = x

Then –8 + 9x – x2 = – (x^{2}– 9x + 8) = – (x^{2}– 8x – x + 8)

= – (x– 8)(x – 1)

= – [(a– b)6– 8][(a – b)6– 1]

= [1 – (a –b)6][(a – b)6– 8]

= [(1)^{3}– {(a – b)^{2}}^{3}][{(a – b)^{2}}^{3}– (2)^{3}]

= [1 – (a –b)^{2}][1 + (a – b)^{4} + (a – b)^{2}][(a – b)^{2}– 2][(a – b)^{4} + 4 + 2(a – b)^{2}] Ans.

**Question **Factorize: 6x^{2}– 5xy – 4y^{2 }+ x + 17y – 15

**Solution: **6x^{2} + x[1 – 5y] – [4y^{2}– 17y + 15]

= 6x^{2} + x[1 – 5y] – [4y^{2}– 17y + 15]

= 6x^{2 }+ x[1 – 5y] – [4y(y – 3) – 5(y – 3)]

= 6x^{2} + x[1 – 5y] – (4y – 5)(y – 3)

= 6x^{2} + 3(y –3)x– 2(4y – 5)x – (4y –)(y - 3)

= 3x[2x + y 3] - (4y - 5)(2x + y - 3)

= (2x + y - 3)(3x - 4y + 5) Ans.