Algebra Chapter 6 Factorisation Exercise 6 Solved Example Problems
Factorisation Introduction
A very important topic in algebra is to resolve an expression into factors. The idea of factors in algebra is similar to that of in arithmetic.
You have already learned about the multiple and factor of a given number in arithmetic.
For example, 35 = 5×7. Hence, factors of 35 are 1, 5, 7, and 35.
In a similar way, the factors of ab are 1, a, b, and ab. Let us now define a factor.
Math Solution Of Class 7 Wbbse
Factor
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If the product of two or more expressions is equal to another expression then those expressions are called the factors of the product.
Example: If p x q x r = x then the expressions p, q, and r are called the factors of x.
Therefore, by factorization of the expression x, three factors p, q, and r are obtained.
Math Solution Of Class 7 Wbbse Factorisation by a selection of common factors
If a polynomial contains one or more common factors in each of its terms then the common factor (or factors) are taken outside the bracket according to the distributive law and the remaining portion is kept inside the bracket.
Example:
⇒ \(a^2 b+a b+a b^2=a b(a+1+b)=a \times b \times(a+1+b)\)
⇒ \(x^3 y^2+x^2 y^3=x^2 y^2(x+y)=x \times x \times y x y x(x+y)\)
Factorization with the help of the formula of the square of a binomial
We know that, (a + b)2 = (a + b)(a + b) and (a -b)2 = (a-b) (a- b).
Some expressions can be factorised with the help of these two formulae.
Example: \(x^2+4 x y z+4 y^2 z^2\)
= \((x)^2+2 \cdot x \cdot 2 y z+(2 y z)^2=(x+2 y z)^2\)
Again, \(4 a^2-12 a b+9 b^2=(2 a)^2-2 \cdot 2 a \cdot 3 b+(3 b)^2\) 4a2 – 12ab + 9b2 = (2a)2 – 2.2a.3b + (3b)2
= \((2 a-3 b)^2=(2 a-3 b)(2 a-3 b)\)
Wbbse Class 7 Maths Solutions
Factorisation with the help of the formula of the difference of two squares
Applying the formula a2-b2 = (a + b) (a-b), some expressions can be factorised.
Example : \(25 x^2-81 y^2=(5 x)^2-(9 y)^2=(5 x+9 y)(5 x-9 y) .\)
Factorisation by the simultaneous application of the formulae of the square of a binomial and the difference of two squares
We know that, a2 + 2ab + b2 = (a + b)2 and a2 -b2 = (a + b) (a-b).
Some expressions may be factorised by applying these two formulae simultaneously.
Example: \(a^4+4\)
= \(\left(a^2\right)^2+(2)^2\)
= \(\left(a^2\right)^2+2 \cdot a^2 \cdot 2+(2)^2-4 a^2\)
= \(\left(a^2+2\right)^2-(2 a)^2=\left(a^2+2+2 a\right)\left(a^2+2-2 a\right)\)
= \(\left(a^2+2 a+2\right)\left(a^2-2 a+2\right)\)
Algebra Chapter 6 Factorisation Exercise 6 Some Examples On Factorisation
Example 1. Factorise : 3x + 12.
Solution:
Given:
3x + 12
= 3.x + 3.4 = 3 (x + 4)
∴ 3 (x +4).
3x + 12 = 3 (x +4).
Example 2. Factorise : 25m – 35n.
Solution:
Given: 25m – 35n
25m – 35n = 5 x 5m – 5 x 7n
∴ 5 (5m- 7n)
25m – 35n = 5 (5m- 7n)
Math Solution Of Class 7 Wbbse
Example 3. Factorise : \(p^2 q r+p q^2 r+p q r^2\)
Solution:
Given: \(p^2 q r+p q^2 r+p q r^2\)
⇒ \(p^2 q r+p q^2 r+p q r^2\) = pqr (p + q + r)
∴pqr (p + q + r).
⇒ \(p^2 q r+p q^2 r+p q r^2\) = pqr (p + q + r).
Wbbse Class 7 Maths Solutions
Example 4. Factorise : \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\)
Solution:
Given: \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\)
⇒ \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\)
∴ \(2 a^2 b\left(2 a^2-3 a b+6 b^2\right)\)
⇒ \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\) = \(2 a^2 b\left(2 a^2-3 a b+6 b^2\right)\)
Example 5. Resolve into factors (ax + by) (p – q) – (ax + by) (q – r).
Solution:
Given:
(ax + by) (p-q)-(ax + by) (q – r)
= (ax +by) {(p-q)-(q- r)}
= (ax + by) (p-q-q + r)
= (ax + by) (p-2q + r)
∴ (ax +by) (p – 2q + r).
(ax + by) (p – q) – (ax + by) (q – r) = (ax +by) (p – 2q + r).
Example 6. Resolve into factors: \(x^2-(a+b) x+a b\)
Solution:
Given:
⇒ \(x^2-(a+b) x+a b\)
⇒ \(x^2-(a+b) x+a b=x^2-a x-b x+a b=x(x-a)-b(x-a)\)
∴ (x-a) (x- b)
⇒ \(x^2-(a+b) x+a b\) = (x-a) (x- b)
Example 7. Resolve into factors : (P + q) (a + b) + r(a + b) + c(p + q + r).
Solution:
Given:
(P + q) (a + b) + r(a + b) + c(p + q + r)
(P + q) (a + b) + r(a + b) + c{p + q + r)
= (a + b) (p + q + r) + c(p + q + r)
∴ (p + q + r) (a + b + c)
(P + q) (a + b) + r(a + b) + c(p + q + r) = (p + q + r) (a + b + c)
Example 8. Factorise : \(\text { 2ay }-x^2-y^2+z^2 \text {. }\)
Solution:
Given:
⇒ \(\text { 2ay }-x^2-y^2+z^2 \text {. }\)
= \((z)^2-(x-y)^2\)
= {z + (x – y)} {z – (x – y)}
= (z + x – y) (z – x + y)
∴ (z + x-y)(z-x + y).
⇒ \(\text { 2ay }-x^2-y^2+z^2 \text {. }\) = (z + x-y)(z-x + y).
Wbbse Class 7 Maths Solutions
Example 9. Resolve into factors \(9(x-y)^2-25(y-z)^2\)
Solution:
Given:
⇒ \(\left.9(x-y)^2-25(y-z)^2=\{3(x-y)\}^2-\{50-z)\right\}^2\)
= \((3 x-3 y)^2-(5 y-5 z)^2\)
= {(3x – 3y) + (5y – 5z)} {(3x – zy) – (5y – 5z)}
= (3x – 3y + 5y – 5z) (3x – 3y – 5y + 5z)
= (3x + 2y – 5z) (3x – 8y + 5z)
∴ (3x + 2y – 5z) (3x -8y+ 5z).
⇒ \(9(x-y)^2-25(y-z)^2\) = (3x + 2y – 5z) (3x -8y+ 5z).
Example 10. Resolve into factors: \(x^4+x^2 y^2+y^4\)
Solution:
Given:
⇒ \(x^4+x^2 y^2+y^4\)
= \(\left(x^2\right)^2+2 \cdot x^2 \cdot y^2+\left(y^2\right)^2-x^2 y^2\)
= \(\left(x^2+y^2\right)^2-(x y)^2\)
= \(\left(x^2+y^2+x y\right)\left(x^2+y^2-x y\right)\)
= \(\left(x^2+x y+y^2\right)\left(x^2-x y+y^2\right)\)
∴ \(\left(x^2+x y+y^2\right)\left(x^2-x y+y^2\right)\)
⇒ \(x^4+x^2 y^2+y^4\) = \(\left(x^2+x y+y^2\right)\left(x^2-x y+y^2\right)\)
Example 11. Resolve into factors : \(x^2-y^2-6 x a+2 y a+8 a^2\)
Solution:
Given:
⇒ \(x^2-y^2-6 x a+2 y a+8 a^2\)
= \(x^2-6 x a+9 a^2-y^2+2 y a-a^2\)
= \((x)^2-2 \cdot x \cdot 3 a+(3 a)^2-\left(y^2-2 y a+a^2\right)\)
= \((x-3 a)^2-(y-a)^2\)
= {(x – 3a) + (y – a)} {(x – 3a) – (y – a)}
= (x – 3a + y – a) (x – 3a – y + a)
= (x + y-4a) (x-y – 2a)
∴ (x + y- 4a) (x-y – 2a).
⇒ \(x^2-y^2-6 x a+2 y a+8 a^2\) = (x + y- 4a) (x-y – 2a).
Wbbse Class 7 Maths Solutions
Example 12. Factorise: \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\)
Solution:
Given:
⇒ \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\)
⇒ \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\)
= \(4 a^2+12 a b+9 b^2-25 c^2-10 c-1\)
= \(\left(4 a^2+12 a b+9 b^2\right)-\left(25 c^2+10 c+1\right)\)
= \(\left\{(2 a)^2+2.2 a \cdot 3 b+(3 b)^2\right\}-\left\{(5 c)^2+2.5 c \cdot 1+(1)^2\right\}\)
= \((2 a+3 b)^2-(5 c+1)^2=\{(2 a+3 b)+(5 c+1)\}\{(2 a+3 b)-(5 c+1)\}\)
= (2a + 3b +5c + 1) (2a + 3b – 5c – 1)
∴ (2a + 3b + 5c + 1) (2a + 3b – 5c – 1).
⇒ \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\) = (2a + 3b + 5c + 1) (2a + 3b – 5c – 1).
Example 13. Factorise: \(9(5 c+6 d)^2-16(3 c-2 d)^2\)
Solution:
Given:
⇒ \(9(5 c+6 d)^2-16(3 c-2 d)^2\)
= \(\{3(5 c+6 d)\}^2-\{4(3 c-2 d)\}^2\)
= \((15 c+18 d)^2-(12 c-8 d)^2\)
= {(15c + 18d) + (12c – 8d)}{(15c + 18d) – (12c – 8d)}
= (15c+18d+12c-8d)(15c+18d-12c+ 8d)
∴ (27c + 10d)(3c + 26d)
⇒ \(9(5 c+6 d)^2-16(3 c-2 d)^2\) = (27c + 10d)(3c + 26d)
Wbbse Class 7 Maths Solutions
Example 14. Factorise : \(x^4+6 x^2 y^2+8 y^4\)
Solution:
Given:
⇒ \(x^4+6 x^2 y^2+8 y^4\)
= \(\left(x^2\right)^2+2 \cdot x^2 \cdot 3 y^2+\left(3 y^2\right)^2-y^4\)
= \(\left.\left(x^2+3 y^2\right)^2-y^2\right)^2\)
= \(\left(x^2+3 y^2+y^2\right)\left(x^2+3 y^2-y^2\right)\)
= \(\left(x^2+4 y^2\right)\left(x^2+2 y^2\right)\)
∴ \(\left(x^2+4 y^2\right)\left(x^2+2 y^2\right)\)
⇒ \(x^4+6 x^2 y^2+8 y^4\) = \(\left(x^2+4 y^2\right)\left(x^2+2 y^2\right)\)
Example 15. Factorise: \(3 x^2-y^2+z^2-2 x y-4 x z\)
Solution:
Given:
⇒ \(3 x^2-y^2+z^2-2 x y-4 x z\)
= \(4 x^2-4 x z+z^2-x^2-2 x y-y^2\)
= \((2 x)^2-2 \cdot 2 x \cdot z+(z)^2-\left(x^2+2 x y+y^2\right)\)
= \((2 x-z)^2-(x+y)^2\)
= {(2x – z) + (x + y)}{(2x – z) – (x + y)}
= (2x – z + x + y){2x – z – x- y)
= (3x + y – z)(x -y-z)
= (x-y- z)(3x + y-z)
∴ (x – y – z){3x + y – z).
⇒ \(3 x^2-y^2+z^2-2 x y-4 x z\) = (x – y – z){3x + y – z).
Example 16. Factorise : \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2\)
Solution:
Given:
⇒ \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2\)
⇒ \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2=(2 b c)^2-\left(b^2+c^2-a^2\right)^2\)
= \(\left\{(2 b c)+\left(b^2+c^2-a^2\right)\right\}\left\{(2 b c)-\left(b^2+c^2-a^2\right)\right\}\)
= \(\left(2 b c+b^2+c^2-a^2\right)\left(2 b c-b^2-c^2+a^2\right)\)
= \(\left\{(b+c)^2-(a)^2\right\}\left\{(a)^2-(b-c)^2\right\}\)
= (b+ c + a)(b + c – a)(a + b – c)(a – b + c)
∴ (a + b + c)(a + b – c)(b + c – a)(c + a – b)
⇒ \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2\) = (a + b + c)(a + b – c)(b + c – a)(c + a – b)
Class Vii Math Solution Wbbse
Example 17. Resolve \(4 a^2-12 a b+9 b^2\) into two linear factors and find their sum.
Solution:
Given:
⇒ \(4 a^2-12 a b+9 b^2\)
= \((2 a)^2-2 \cdot 2 a \cdot 3 b+(3 b)^2\)
= \((2 a-3 b)^2=(2 a-3 b)(2 a-3 b)\)
Now, sum of the linear factors = 2a – 3b + 2a – 3b
∴ 4a – 6b
⇒ \(4 a^2-12 a b+9 b^2\) = 4a – 6b
Example 18. Resolve \(\frac{p}{q}\) – \(\frac{q}{p}\) into factors.
Solution:
⇒ \(\frac{p}{q}\) – \(\frac{q}{p}\)
= \(\frac{p^2-q^2}{p q}=\frac{(p+q)(p-q)}{p q}\)
= \(\left(\frac{p+q}{p}\right)\left(\frac{p-q}{q}\right)=\left(1+\frac{q}{p}\right)\left(\frac{p}{q}-1\right)\)
⇒ \(\frac{p}{q}\) – \(\frac{q}{p}\) = \(\left(\frac{p+q}{p}\right)\left(\frac{p-q}{q}\right)=\left(1+\frac{q}{p}\right)\left(\frac{p}{q}-1\right)\)
Example 19. Express a – 6 as the difference of two squares and factories.
Math Solution Of Class 7 Wbbse
Solution:
a – b
= \(\left(a^{\frac{1}{2}}\right)^2-\left(b^{\frac{1}{2}}\right)^2=\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)
Example 20. Three factors of an expression are a, a + \(\frac{1}{a}\) and a – \(\frac{1}{a}\) find the expression.
Solution:
The required express
= \(a\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right)=a\left(a^2-\frac{1}{a^2}\right)=a^3-\frac{1}{a}\)
Example 21. Factorise : \(x^4-3 x^2 y^2+9 y^4\)
Solution:
Given:
⇒ \(x^4-3 x^2 y^2+9 y^4\)
= \(\left(x^2\right)^2+2 \cdot x^2 \cdot 3 y^2+\left(3 y^2\right)^2-9 x^2 y^2\)
= \(\left(x^2+3 y^2\right)^2-(3 x y)^2=\left(x^2+3 y^2+3 x y\right)\left(x^2+3 y^2-3 x y\right)\)
∴ \(\left(x^2+3 y^2+3 x y\right)\left(x^2+3 y^2-3 x y\right)\)
⇒ \(x^4-3 x^2 y^2+9 y^4\) = \(\left(x^2+3 y^2+3 x y\right)\left(x^2+3 y^2-3 x y\right)\)
Class Vii Math Solution Wbbse
Example 22. Factorise: \(\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d\)
Solution:
Given:
⇒ \(\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d\)
= \(a^2 c^2-a^2 d^2-b^2 c^2+b^2 d^2-4 a b c d\)
= \(a^2 c^2+b^2 d^2-2 a b c d-a^2 d^2-b^2 c^2-2 a b c d\)
= \(\left(a^2 c^2+b^2 d^2-2 a b c d\right)-\left(a^2 d^2+b^2 c^2+2 a b c d\right)\)
= \((a c-b d)^2-(a d+b c)^2\)
= (ac -bd + ad + bc) (ac – bd – ad – bc)
∴ (ac-bd + ad + bc) (ac-bd-ad-bc)
∴ \(\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d\) = (ac-bd + ad + bc) (ac-bd-ad-bc)
Math Solution Of Class 7 Wbbse
Example 23. Resolve into factors \(3 x^2-y^2-z^2-2 x y-4 x z\)
Solution:
Given:
⇒ \(3 x^2-y^2-z^2-2 x y-4 x z\)
= \(4 x^2-x^2-y^2+z^2-2 x y-4 x z=4 x^2+z^2-4 x z-x^2-y^2-2 x y\)
= \((2 x-z)^2-(x+y)^2=(2 x-z+x+y)(2 x-z-x-y)\)
= (3x+y-z) (x-y-z)
∴ (3x + y-z) (x-y – z).
∴ \(3 x^2-y^2-z^2-2 x y-4 x z\) = (3x + y-z) (x-y – z).
Example 24. Factorise : 6xy – 9y + 4x – 6
Solution:
Given:
6xy – 9y + 4x – 6
6xy – 9y + 4x – 6 = 3y (2x – 3) + 2 (2x – 3)
∴ (2x – 3) ( 3y + 2)
6xy – 9y + 4x – 6 = (2x – 3) ( 3y + 2)
Example 25. Factorise: \(3 x^4+2 x^2 y^2-y^4\)
Solution:
Given:
⇒ \(3 x^4+2 x^2 y^2-y^4\)
= \(3 x^4+3 x^2 y^2-x^2 y^2-y^4\)
= \(3 x^2\left(x^2+y^2\right)-y^2\left(x^2+y^2\right)\)
∴ \(\left(x^2+y^2\right)\left(3 x^2-y^2\right)\)
∴ \(3 x^4+2 x^2 y^2-y^4\) = \(\left(x^2+y^2\right)\left(3 x^2-y^2\right)\)