## Factorise A Polynomial By Splitting The Middle Term Example Problems With Solutions

**Type I: Factorization of Quadratic polynomials of the form x ^{2} + bx + c.**

(i) In order to factorize x

^{2}+ bx + c we have to find numbers p and q such that p + q = b and pq = c.

(ii) After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

**Example 1: **Factorize each of the following expressions:

(i) x^{2} + 6x + 8 (ii) x^{2} + 4x –21

**Solution:**

(i) In order to factorize x^{2} + 6x + 8, we find two numbers p and q such that p + q = 6 and pq = 8.

Clearly, 2 + 4 = 6 and 2 × 4 = 8.

We know split the middle term 6x in the given quadratic as 2x + 4x, so that

x^{2} + 6x + 8 = x^{2} + 2x + 4x + 8

= (x^{2} + 2x) + (4x + 8)

= x (x + 2) + 4 (x+ 2)

= (x + 2) (x + 4)

(ii) In order to factorize x^{2} + 4x – 21, we have to find two numbers p and q such that

p + q = 4 and pq = – 21

Clearly, 7 + (– 3) = 4 and 7 × – 3 = – 21

We now split the middle term 4x of

x^{2} + 4x – 21 as 7x – 3x, so that

x^{2} + 4x – 21 = x^{2} + 7x – 3 x – 21

= (x^{2} + 7x) – (3x + 21)

= x (x + 7) – 3 (x + 7) = (x + 7) (x – 3)

**Example 2: **Factorize each of the following quadratic polynomials: x^{2} – 21x + 108

**Solution: **In order to factorize x^{2} – 21x + 108,

we have to find two numbers such that their sum is – 21 and the product 108.

Clearly, – 21 = – 12– 9 and – 12 × – 9 = 108

x^{2} – 21 x + 108 = x^{2} – 12 x – 9x + 108

= (x^{2} – 12 x) – (9x– 108)

= x(x – 12) – 9 (x – 12) = (x–12) (x – 9)

**Example 3: **Factorize the following by splitting the middle term : x^{2} + 3 √3 x + 6

**Solution: **In order to factorize x^{2} + 3 √3 x + 6, we have to find two numbers p and q such that

**Type II: Factorization of polynomials reducible to the form x ^{2} + bx + c.**

**Example 4: **Factorize (a^{2} – 2a)^{2} – 23(a^{2} – 2a) + 120.

**Solution:**

**Example 5: **Factorize the following by splitting the middle term x^{4}– 5x^{2} + 4

**Solution:**

Let x^{2} = y. Then, x^{4} – 5x^{2} + 4

= y^{2} – 5 y + 4

Now, y^{2} – 5 y + 4

= y^{2} – 4y – y + 4

= (y^{2} – 4y) – (y – 4)

= y(y –4) – (y– 4)

= (y – 4) (y – 1)

Replacing y by x^{2} on both sides, we get

x^{4} – 5x^{2} + 4 = (x^{2}–4) (x^{2} – 1)

= (x^{2}–2^{2}) (x^{2} – 1^{2}) = (x–2) (x+2) (x – 1) (x + 1)

**Example 6: **Factorize (x^{2} – 4x) (x^{2} – 4x – 1) – 20

**Solution: **The given expression is

(x

^{2}– 4x) (x

^{2}– 4x – 1) – 20

= (x

^{2}– 4x)

^{2}– (x

^{2}– 4x) – 20

Let x

^{2}– 4x = y . Then,

(x

^{2}– 4x)

^{2}– (x

^{2}– 4x) – 20 = y

^{2}– y – 20

Now, y

^{2}– y – 20

= y

^{2}–5 y + 4y – 20

= (y

^{2}– 5 y) + (4y– 20)

= y (y – 5) + 4 (y – 5)

= (y – 5) (y + 4)

Thus, y

^{2}– y – 20 = (y – 5) (y + 4)

Replacing y by x

^{2}– 4x on both sides, we get

(x

^{2}– 4x)

^{2}– (x

^{2}– 4x) – 20

= (x

^{2}– 4x – 5) (x

^{2}– 4x +4)

= (x

^{2}– 5x + x – 5) (x

^{2}– 2 × x × 2 + 2

^{2})

= {x (x – 5) + (x – 5)} (x – 2)

^{2}

= (x – 5) (x + 1) (x – 2)

^{2}

**Type III: Factorization of Expressions which are not quadratic but can factorized by splitting the middle term.**

**Example 7: **If x^{2} + px + q = (x + a) (x + b), then factorize x^{2} + pxy + qy^{2}.

**Solution: **We have,

x^{2} + px + q = (x + a) (x + b)

⇒ x^{2} + px + q = x^{2} + x(a + b) + ab

On equating the coefficients of like powers of x, we get

p = a + b and q = ab

∴ x^{2} + pxy + qy^{2} = x^{2} + (a + b)xy + aby^{2}

= (x^{2} + axy) + (bxy + aby^{2})

= x(x + ay) + by(x + ay)

= (x + ay) (x + by)

**Example 8: **Factorize the following expression x^{2}y^{2} – xy – 72

**Solution:**

In order to factorize x^{2}y^{2} – xy – 72, we have to find two numbers p and q such that

p+ q = – 1 and pq = – 72

clearly, – 9 +8 = – 1 and – 9 × 8 = – 72.

So, we write the middle term – xy of

x^{2}y^{2} – xy – 72 as – 9 xy + 8 xy, so that

x^{2}y^{2} – xy – 72 = x^{2}y^{2} – 9 xy + 8 xy – 72

= (x^{2}y2^{2} – 9xy) + (8xy – 72)

= xy (xy – 9) + 8 (xy – 9)

= (xy – 9) (xy + 8)

**Factorization Of Polynomials Of The Form ****ax**^{2} + bx + c, a ≠** ****0, 1**

^{2}+ bx + c, a ≠

** Type I: Factorization of quadratic polynomials of the form ax ^{2} + bx + c, a 0, 1**

(i) In order to factorize ax

^{2}+ bx + c. We find numbers l and m such that l + m = b and lm = ac

(ii) After finding l and m, we split the middle term bx as lx + mx and get the desired factors by grouping the terms.

**Example 9: **Factorize the following expression

6x^{2} – 5 x – 6

**Solution: **The given expression is of the form ax^{2}+ bx+c, where, a = 6, b = – 5 and c = –6.

In order to factorize the given expression, we have to find two numbers l and m such that

l + m = b = i.e., l + m = – 5

and lm = ac i.e. lm = 6 × – 6 = – 36

i.e., we have to find two factors of – 36

such that their sum is – 5. Clearly,

– 9 + 4 = – 5 and – 9 × 4 = – 36

l = – 9 and m = 4

Now, we split the middle term – 5x of

x^{2} – 5x – 6 as – 9 x + 4x, so that

6x^{2} – 5x – 6 = 6x^{2}–9x + 4x – 6

= (6x^{2} – 9x) + (4x – 6)

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

**Example 10: **Factorize each of the following expressions:

(i) √3 x^{2} + 11x + 6 √3

(ii) 4 √3 x^{2} + 5x – 2 √3

(iii) 7 √2 x^{2} – 10 x – 4 √2

**Solution: **(i) The given quadratic expression is of the form ax^{2} + bx + c,

where a = √3, b = 11 and c = 6 √3.

In order to factorize it, we have to find two numbers l and m such that

**Example 11: **Factorize the following by splitting the middle term

1/3 x^{2} – 2x – 9

**Solution: **

**Type II: Factorization of trinomial expressions which are not quadratic but can be factorized by splitting the middle term.**

**Example 12: **Factorize the following trinomial by splitting the middle term

8a^{3} – 2a^{2}b – 15 ab^{2}

**Solution: **Here a^{3} × ab^{2} = (a^{2}b)^{2} i.e., the product of the variables in first and last term is same as the square of the variables in the middle term. So, in order to factorize the given trinomial, we split the middle term

– 2a^{2}b as – 12a^{2}b + 10 a^{2}b , so that

8a^{3} – 2a^{2}b – 15 ab^{2}

= 8a^{3} –12a^{2}b +10 a^{2}b–15 ab^{2}

= 4a^{2}(2a – 3b) + 5 ab (2a – 3b)

= (2a – 3b) (4a^{2} + 5ab)

= (2a – 3b) a (4a + 5b)

= a (2a – 3 b) (4a + 5b)

**Type III : Factorization of trinomial expressions reducible to quadratic expressions.**

**Example 13: **Factorize each of the following expressions by splitting the middle term :

(i) 9(x – 2y)^{2}– 4(x – 2y) – 13

(ii) 2(x + y)^{2} – 9(x + y) – 5

(iii) 8(a + 1)^{2} + 2(a + 1) (b + 2) – 15(b + 2)^{2}

**Solution: ** (i) The given expression is 9(x – 2y)^{2} – 4(x – 2y) – 13.

Putting x – 2y = a, we get

9(x – 2y)^{2} – 4(x – 2y) – 13 = 9a^{2} – 4a – 13

Now, 9a^{2} – 4a – 13 = 9a^{2} – 13a + 9a – 13

= (9a^{2} – 13a) + (9a – 13)

= a(9a – 13) + (9a – 13)

= (a + 1) (9a – 13)

Replacing a by x – 2y on both sides, we get

9(x – 2y)^{2} – 4(x – 2y) – 13

= (x – 2y + 1) {9(x – 2y) – 13}

= (x – 2y + 1) (9x – 18y – 13)

(ii) The given expression is

2(x + y)^{2} – 9(x + y) – 5

Replacing x + y by a in the given expression, we have

2(x + y)^{2} – 9(x + y) – 5 = 2a^{2} – 9a – 5

Now, 2a^{2} – 9a – 5 = 2a^{2} – 10a + a – 5

= (2a^{2} – 10a) + (a – 5)

= 2a(a – 5) + (a – 5) = (a – 5) (2a + 1)

Replacing a by x + y on both sides, we get

2(x + y)^{2} – 9(x + y) – 5

= (x + y – 5) {2(x + y) + 1}

= (x + y – 5) (2x + 2y + 1).

(iii) The given trinomial is

8(a + 1)^{2} + 2(a + 1) (b + 2) – 15(b + 2)^{2}

Putting a + 1 = x and b + 2 = y, we have

8(a + 1)^{2} + 2(a + 1) (b + 2) – 15(b + 2)^{2}

= 8x^{2} + 2xy – 15y^{2}

= 8x^{2} + 12xy – 10xy – 15y^{2}

= 4x(2x + 3y) – 5y(2x + 3y)

= (2x + 3y) (4x – 5y)

Replacing x by a + 1 and y by b + 2, we get

8(a + 1)^{2} + 2(a + 1) (b + 2) – 15(b + 2)^{2}

= {2(a + 1) + 3(b + 2)} {4(a + 1) – 5(b +2)}

= (2a + 3b + 8) (4a – 5b – 6)