Question

Ten examples of middle term splitting.

ONLY FOR CLASS 9 TO 12​

Answer 1

Step-by-step explanation:

here is the 10 examples of middle term splitting...

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Answer 2

Example 1: Factorize each of the following expressions:

(i) x2 + 6x + 8 (ii) x2 + 4x –21

Solution:

(i) In order to factorize x2 + 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

x2 + 6x + 8 = x2 + 2x + 4x + 8

= (x2 + 2x) + (4x + 8)

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

= (x + 2) (x + 4)

(ii) In order to factorize x2 + 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

x2 + 4x – 21 as 7x – 3x, so that

x2 + 4x – 21 = x2 + 7x – 3 x – 21

= (x2 + 7x) – (3x + 21)

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

Example 2: Factorize each of the following quadratic polynomials: x2 – 21x + 108

Solution: In order to factorize x2 – 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

x2 – 21 x + 108 = x2 – 12 x – 9x + 108

= (x2 – 12 x) – (9x– 108)

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

Example 3: Factorize the following by splitting the middle term : x2 + 3 √3 x + 6

Solution: In order to factorize x2 + 3 √3 x + 6, we have to find two numbers p and q such that

How To Factorise A Polynomial By Splitting The Middle Term 1

Type II: Factorization of polynomials reducible to the form x2 + bx + c.

Example 4: Factorize (a2 – 2a)2 – 23(a2 – 2a) + 120.

Solution:

How To Factorise A Polynomial By Splitting The Middle Term 2

Example 5: Factorize the following by splitting the middle term x4– 5x2 + 4

Solution:

Let x2 = y. Then, x4 – 5x2 + 4

= y2 – 5 y + 4

Now, y2 – 5 y + 4

= y2 – 4y – y + 4

= (y2 – 4y) – (y – 4)

= y(y –4) – (y– 4)

= (y – 4) (y – 1)

Replacing y by x2 on both sides, we get

x4 – 5x2 + 4 = (x2–4) (x2 – 1)

= (x2–22) (x2 – 12) = (x–2) (x+2) (x – 1) (x + 1)

Example 6: Factorize (x2 – 4x) (x2 – 4x – 1) – 20

Solution:

The given expression is

(x2 – 4x) (x2 – 4x – 1) – 20

= (x2 – 4x)2 – (x2 – 4x) – 20

Let x2 – 4x = y . Then,

(x2 – 4x)2 – (x2 – 4x) – 20 = y2 – y – 20

Now, y2 – y – 20

= y2 –5 y + 4y – 20

= (y2 – 5 y) + (4y– 20)

= y (y – 5) + 4 (y – 5)

= (y – 5) (y + 4)

Thus, y2 – y – 20 = (y – 5) (y + 4)

Replacing y by x2 – 4x on both sides, we get

(x2 – 4x)2 – (x2 – 4x) – 20

= (x2 – 4x – 5) (x2 – 4x +4)

= (x2 – 5x + x – 5) (x2 – 2 × x × 2 + 22)

= {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 x2 + px + q = (x + a) (x + b), then factorize x2 + pxy + qy2.

Solution: We have,

x2 + px + q = (x + a) (x + b)

⇒ x2 + px + q = x2 + x(a + b) + ab

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

p = a + b and q = ab

∴ x2 + pxy + qy2 = x2 + (a + b)xy + aby2

= (x2 + axy) + (bxy + aby2)

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

= (x + ay) (x + by)

Example 8: Factorize the following expression x2y2 – xy – 72

Solution:

In order to factorize x2y2 – 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

x2y2 – xy – 72 as – 9 xy + 8 xy, so that

x2y2 – xy – 72 = x2y2 – 9 xy + 8 xy – 72

= (x2y22 – 9xy) + (8xy – 72)

= xy (xy – 9) + 8 (xy – 9)

= (xy – 9) (xy + 8)