Algebra Chapter 7 Factorisation By Breaking The Middle Term
Factorization By Breaking The Middle-Term Introduction
In the previous chapter, we resolved the algebraic expressions into factors by using the formulae. Here also we aim to resolve the algebraic expressions into factors by using a different method— the method of breaking the middle term. Although this method is easier than the method discussed in the previous chapter, we cannot apply this method to all algebraic expressions due to some restrictions. This method applies only to those algebraic expressions which have exactly three terms and the power of the variable of the first term is twice that of the second when arranged in descending order of the power.
Factorization of the expressions of the form x² + px + q
Any algebraic expression of the form \(x^2+p x+q\) is called a quadratic expression since the highest power of x is 2. This type of expression can be factorized by splitting the middle coefficients. Here we should find two quantities such that their algebraic sum is p and product is q Let us consider the expression,
x + (a + b)x + ab.
It is of the form \(x^2+p x+q\), where a + b -p and ab = q.
Now, \(x^2+(a+b) x+a b\)
= \(x^2+a x+b x+a b\)
= x (x + a) + b(x + a)
= (x + a) (x + b).
So we may conclude that x2 + px + q can be resolved into two linear factors (x + a) and (x + 6), when a + b = p and ab = q.
Read And Learn More WBBSE Solutions For Class 8 Maths
Algebra Chapter 7 Factorization By Breaking The Middle Term Some Examples
Example 1
Factorize : \(x^2+7 x+12\)
Solution :
Given: \(x^2+7 x+12\)
\(x^2+7 x+12\)= \(x^2\) + (4 + 3)x + 12
= \(x^2\) + 4x + 3x + 12
= x (x + 4) + 3(x + 4)
= (x + 4) (x + 3)
\(x^2\) + 7x + 12 = \(x^2\) + (4 + 3)x + 12
WBBSE Class 8 Factorisation by Middle Term Notes
Example 2
Factorize : \(x^2-3 x-10\).
Solution :
Given: \(x^2-3 x-10\)
\(x^2-3 x-10\)= \(x^2 – (5 – 2)x – 10\)
= \(x^2 – 5x + 2x – 10\)
= x(x – 5) + 2(x – 5)
= (x – 5) (x + 2)
\(x^2-3 x-10\) = (x – 5) (x + 2)
Example 3
Factorize : \(x^2-9 x+20\)
Solution :
Given: \(x^2-9 x+20\)
\(x^2-9 x+20\)= \(x^2 – (5 + 4)x + 20\)
= \(x^2 – 5x – 4x + 20\)
= x(x – 5) – 4(x – 5)
= (x – 5) (x – 4)
\(x^2 – 9x + 20\) = (x – 5) (x – 4)
Example 4
Factorize : \(x^2+2 x-35\)
Solution :
Given \(x^2+2 x-35\)
\(x^2+2 x-35\)= \(x^2+(7-5) x-35 s=x^2+7 x-5 x-35\) = x(x + 7) – 5(x + 7)
= (x + 7) (x – 5)
\(x^2+2 x-35\) = (x + 7) (x – 5)
Example 5
Factorize :
Solution:
Given: \(x^2+20 x y+75 y^2\)
\(x^2+20 x y+75 y^2\)= \(x^2+(15+5) x y+75 y^2\)
= \(x^2+15 x y+5 r y+75 y^2\)
= x(x + 15y) + 5y(x + 15y)
= (x + 15y) (x + by)
\(x^2+20 x y+75 y^2\). = (x + 15y) (x + by)
Understanding Middle Term Factorisation
Example 6
Factorize : \((x+y)^2-11(x+y)+30\)
Solution :
Given: \((x+y)^2-11(x+y)+30\)
\((x+y)^2-11(x+y)+30\)= \((x+y)^2\) – 6(x + y) – 5(x + y) + 30
= (x + y) (x + y – 6) – 5(x + y – 6)
= (x + y- 6) (x +y- 5)
\((x+y)^2-11(x+y)+30\) = (x + y- 6) (x +y- 5)
Example 7
Factorize : \((a+b)^2+(a+b)-56\)
Solution:
Given: \((a+b)^2+(a+b)-56\)
\((a+b)^2+(a+b)-56\)= (a + b)2 + 8(a + b) – 7(a + b) – 56
= (a + b) (a + b + 8) – 7(a + 6 + 8)
= (a + b + 8) (a + b – 7)
\((a+b)^2+(a+b)-56\) = (a + b + 8) (a + b – 7)
Example 8
Factorize : \(x^2+x-(a+1)(a+2)\)
Solution :
Given: \(x^2+x-(a+1)(a+2)\)
\(x^2+x-(a+1)(a+2)\)= \(x^2\) + {(a + 2) – (a + 1)}x – (a + 1)(a + 2)
= \(x^2\) + (a + 2)x – (a + 1)x – (a + 1) (a + 2)
= x(x + a + 2) – (a + 1) (x + a + 2)
= (x + a + 2) (x – a – 1)
\(x^2\)+x-(a+1)(a+2) = (x + a + 2) (x – a – 1)
Step-by-Step Guide to Breaking the Middle Term
Example 9
Factorize : \(x^4-13 x^2+42\)
Solution :
Given: \(x^4-13 x^2+42\)
\(x^4-13 x^2+42\)= \(x^4-(6+7) x^2+42\)
= \(x^4-6 x^2-7 x^2+42\)
= \(x^2\left(x^2-6\right)-7\left(x^2-6\right)\)
= \(\left(x^2-6\right)\left(x^2-7\right)\)
\(x^4-13 x^2+42=\left(x^2-6\right)\left(x^2-7\right)\)Example 10
Factorize : \(x^6-10 x^3+16\)
Solution :
Given:
\(x^6-10 x^3+16\)= \(x^6-(8+2) x^3+16\)
= \(x^6-8 x^3-2 x^3+16\)
= \(x^3\left(x^3-8\right)-2\left(x^3-8\right)\)
= \(\left(x^3-8\right)\left(x^3-2\right)\)
= \(\left\{(x)^3-(2)^3\right\}\left(x^3-2\right)\)
= \((x-2)\left(x^2+2 x+4\right)\left(x^3-2\right)\)
\(x^6-10 x^3+16=(x-2)\left(x^2+2 x+4\right)\left(x^3-2\right)\)Example 11
Factorize : (p – 1) (p – 2) (p + 3) (p + 4) + 6.
Solution:
Given
(p – 1) (p – 2) (p + 3) (p + 4) + 6
(p – 1) (p – 2) (p + 3) (p + 4) + 6
= {(p – 1) (p + 3} {(p – 2) (p + 4)} + 6
= (p2 + 3p – p – 3) (p2 + 4p – 2p – 8) + 6
= (p2 + 2p – 3) (p2 + 2p – 8) + 6
Let, p2 + 2p = a.
Then the given expression
= (a-3) (a-8)+ 6 = a2 – 8a – 3a + 24 + 6
= a2 – 11a + 30 = a2 – (6 + 5)a + 30
= a2 – 6a – 5a + 30
= a(a – 6) – 5(a – 6)
= (a – 6) (a – 5)
= (p2 + 2p – 6) (p2 + 2p – 5) [∵ putting the value of a]
(p – 1) (p – 2) (p + 3) (p + 4) + 6 = (p2 + 2p – 6) (p2 + 2p – 5)
Practice Problems on Middle Term Factorisation
Example 12
Factorize : p2 + ( a+1/a)p + 1.
Solution:
Given p2 + ( a+1/a)p + 1.
p2 – (a+1/a)p + 1
=p2 – a . p – 1/a . p + 1/a . a
= p(p-a) – 1/a (p-a)
= (p-a)(p – 1/a)
p2 + ( a+1/a)p + 1. = (p-a)(p – 1/a)
Example 13
Factorize : \(x^2\) – bx – (a + 36)(a + 26).
Solution :
Given \(x^2\) – bx – (a + 36)(a + 26).
\(x^2\) – bx – (a + 3b)(a + 2b)
= \(x^2\) – {(a + 3b) – (a + 2b)}x – (a + 3b)(a + 2b)
= \(x^2\) – (a + 3b)x +(a + 2b)x-(a+3b)(a+2b)
= x{x – a – 3b} + (a + 2b){x – a – 3b}
= (x – a – 3b)(x + a + 2b)
\(x^2\) – bx – (a + 36)(a + 26). = (x – a – 3b)(x + a + 2b)
Example 14
Factorize : \((x+y)^2-5 x-5 y+6\)
Solution :
Given \((x+y)^2-5 x-5 y+6\)
\((x+y)^2-5 x-5 y+6\)= \((x+y)^2-5 x-5 y+6\)
Let, x + y = a.
Then the given expression
= \(a^2-5 a+6=a^2-2 a-3 a+6\)
= a(a – 2) – 3 (a – 2) = (a – 2) (a – 3)
= (x + y-2)(x + y-3) [∵ putting the value of a ]
\((x+y)^2-5 x-5 y+6\) = (x + y-2)(x + y-3)
Example 15
Factorize : (x + 1)(x + 3)(x – 4)(x – 6) + 24.
Solution :
Given (x + 1)(x + 3)(x – 4)(x – 6) + 24
= (x + l)(x – 4)(x + 3)(x – 6) + 24
= (x2 – 4x + x – 4XX2 – 6x + 3x – 18) + 24
= (x2 – 3x – 4)(x2 – 3x – 18) + 24
Let, x2 – 3x = a.
Then the given expression
= a2 – 18a – 4a + 72 + 24
= a2 – 22a + 96
= a2 – 6a – 16a + 96
= a(a – 6) — 16(a — 6)
= (a – 6) (a – 16)
= (x2-3x-6)(x2-3x- 16) [∵ putting the value of a]
(x + 1)(x + 3)(x – 4)(x – 6) + 24 = (x2-3x-6)(x2-3x- 16)
Factorization of the expressions of the form ax 2+ bx +c
This type of quadratic expression can also be factorized by splitting the middle coefficients. Here we should find two quantities such that their algebraic sum is b and the product is ac.
Algebra Chapter 7 Factorization By Breaking The Middle Term Some Examples
Example 1
Factorize : 3x2 + 8x + 5.
Solution :
Given: 3X2 + 8x + 5
3X2 + 8x + 5
= 3X2 + (3 + 5)x + 5
= 3x2 + 3x + 5x + 5
= 3x(x + 1) + 5(x + 1)
= (x + 1) (3x + 5)
3X2 + 8x + 5 = (x + 1) (3x + 5)
Examples of Quadratic Factorisation by Middle Term
Example 2
Factorize : 6x2 + x – 40.
Solution:
Given 6x2 + x – 40.
6x2 + x – 40
= 6x2 + (16 – 15)x – 40
= 6x2 + 16x – 15* – 40
= 2x(3x + 8) – 5(3x + 8)
= (3x + 8) (2x – 5)
6x2 + x – 40. = (3x + 8) (2x – 5)
Example 3
Factorize : 10x2 – x – 24.
Solution :
Given 10x2 – x – 24.
10x2 – x – 24
= 10x2 + (- 16 + 15)x – 24
= 10x2 – I6x + 15x – 24
= 2x(5x – 8) + 3(5x – 8)
= (5x – 8) (2x + 3)
10x2 – x – 24. = (5x – 8) (2x + 3)
Example 4
Factorize : 63x – 59x + 10.
Solution :
Given 63x – 59x + 10.
63x2 – 59x + 10
= 63x2 + (- 45 – 14)x + 10
= 63x2 – 45x – 14x + 10
= 9x (7x – 5) – 2(7x – 5)
= (7x – 5) (9x – 2)
63x2 – 59x + 10 = (7x – 5) (9x – 2)
Example 5
Factorize : 10x2 + 31xy + 24y2.
Solution :
Given 10x2 + 31xy + 24y2.
10x2 + 31xy + 24y2
= 10x2 + (15 + 16)xy + 24y2
= 10x2 + 15xy + 16xy + 24y2
= 5x(2x + 3y) + 8y (2x + 3y)
= (2x + 3y) (5x + 8y)
10x2 + 31xy + 24y2. = (2x + 3y) (5x + 8y)
Conceptual Questions on Quadratic Factorisation
Example 6
Factorize : 21 (a + b)2 + 8(a + b) – 45.
Solution :
Given 21 (a + b)2 + 8(a + b) – 45.
21(a + b)2 + 8(a + 6) – 45
= 21(a + b)2 + 35(a + b) – 27(a + b) – 45
= 7(a + b) {3(a + b) + 5} – 9 {3(a + b) + 5}
= {3(a + b) + 5} (7(a + b) – 9}
= (3a + 35 + 5) (7a + 75-9)
21 (a + b)2 + 8(a + b) – 45. = (3a + 35 + 5) (7a + 75-9)
Example 7
Factorize : 15(x + y)2 – 26(x + y) + 8.
Solution :
Given 15(x + y)2 – 26(x + y) + 8.
15(x + y)2 – 26(x + y) + 8
= 15(x + y)2 – 20(x + y) – 6(x + y) + 8
= 5(x + y) {3(x + y) – 4} -2{3(x + y) – 4}
= {3(x + y) – 4} {5(x + y) – 2}
= (3x + 3y-4)(5x + 5y-2)
15(x + y)2 – 26(x + y) + 8. = (3x + 3y-4)(5x + 5y-2)
Example 8
Factorize : (a – 1)x2 + a2xy + (a + 1)y2.
Solution :
Given (a – 1)x2 + a2xy + (a + 1)y2.
(a – 1)x2 + a2xy + (a + 1)y2
= (a – 1)x2 + {(a2 — 1) + l}xy + (a + 1)y2
= (a – 1)x2 + (a2 – 1):xy + xy + (a + 1)y2
= (a – 1)x {x + (a + l)y} + y{x + (a + 1)y}
= {x + (a + 1)y} {(a – 1)x + y}
= (x + ay + y) (ax-x +y)
(a – 1)x2 + a2xy + (a + 1)y2. = (x + ay + y) (ax-x +y)
Example 9
Factorize : 6x4 + 17x2 – 45.
Solution:
Given 6x4 + 17x2 – 45.
6x4 + 17x2 – 45
= 6x4 + (27 – 10)x2 – 45
= 6x4 + 27x2 – 10x2 – 45
= 3x2(2x2 + 9) – 5(2x2 + 9)
= (2x2 + 9) (3x2 – 5)
6x4 + 17x2 – 45. = (2x2 + 9) (3x2 – 5)
Example 10
Factorize : 21x6 — 29x3 +10.
Solution :
Given 21x6 — 29x3 +10.
2 1x6 – 29x3 + 10
= 21x6 – (14 + 15)x3 + 10
= 2 1x6 – 14x3 – 15x3 + 10
= 7x3(3x3 – 2) – 5(3x3 – 2)
= (3x3 – 2) (7x3 — 5)
21x6 — 29x3 +10. = (3x3 – 2) (7x3 — 5)
Example 11
Factorize : 5{a2-b2)2— 8ab (a2 – b2) – 13a2b2.
Solution :
Given 5{a2-b2)2— 8ab (a2 – b2) – 13a2b2.
Let a – b = x and ab = -y
To the given expression
= 5x2 – 8xy – 13y2
= 5x2 – (13 – 5)xy – 13y2
= 5x2 – 13xy + 5xy – 13y2
= x(5x – 13y) + y(5x – 13;y)
= (5x – 13y) (x + y)
= {5(a2 – b2) – 13ab} {a2 – b2 + ab}
= (5a2 – 5b2 – 13a6) (a2 – b2 + ab)
5{a2-b2)2– 8ab (a2 – b2) – 13a2b2. = (5a2 – 5b2 – 13ab) (a2 – b2 + ab)
Example 12
Factorize : (x + 1) (x + 2) (3x— 1)(3x – 4) + 12.
Solution :
Given (x + 1) (x + 2) (3x— 1)(3x – 4) + 12.
(x + 1) (x + 2) (3x-l) (3x – 4) + 12
= {(x + 1) (3x – 1)} {(x + 2) (3x – 4)} + 12
= (3x2 – x + 3x – 1) (3x2 – 4x + 6x – 8) + 12
= (3x2 + 2x- 1) (3x2 + 2x – 8) + 12 Let, 3x2 + 2x = a
The given expression
= (a – 1) (a – 8) + 12
= a2 – 8a – a + 8 + 12
= a2 – 9a + 20
= a – 5a – 4a + 20
= a(a -5)- 4 (a – 5)
= (a – 5) (a – 4)
= (3x2 + 2x – 5) (3x2 + 2x – 4)
= (3x2 + 5x – 3x – 5) (3x2 + 2x – 4)
= {x(3x + 5) -1(3x + 5)} (3x2 + 2x – 4)
= (3x + 5) (x- 1) (3x2 + 2x- 4)
(x + 1) (x + 2) (3x – 1)(3x – 4) + 12. = (3x + 5) (x – 1) (3x2 + 2x – 4)