Class 9
Chapter 2 POLYNOMIAL
Ten examples of middle term splitting.
ONLY FOR CLASS 9 TO 12
Answers
Answer:
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)
Factorization Of Polynomials Of The Form ax2 + bx + c, a ≠ 0, 1
Type I: Factorization of quadratic polynomials of the form ax2 + bx + c, a 0, 1
(i) In order to factorize ax2 + 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
6x2 – 5 x – 6
Solution: The given expression is of the form ax2+ 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
x2 – 5x – 6 as – 9 x + 4x, so that
6x2 – 5x – 6 = 6x2–9x + 4x – 6
= (6x2 – 9x) + (4x – 6)
= 3x (2x – 3) + 2(2x – 3) = (2x – 3) (3x + 2)
Example 10: Factorize each of the following expressions:
(i) √3 x2 + 11x + 6 √3
(ii) 4 √3 x2 + 5x – 2 √3
(iii) 7 √2 x2 – 10 x – 4 √2
Solution: (i) The given quadratic expression is of the form ax2 + 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
How To Factorise A Polynomial By Splitting The Middle Term 3
How To Factorise A Polynomial By Splitting The Middle Term 4
How To Factorise A Polynomial By Splitting The Middle Term 5
How To Factorise A Polynomial By Splitting The Middle Term 6
Step-by-step explanation:
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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)