Factorisation problems are provided here with solutions for students. Factorisation questions and solutions for students of Class 7, Class 8, Class 9 and Class 10 are given to make them practise algebra and polynomial concepts. The factorisation is a method of factoring a number or a polynomial. The polynomials are decomposed into products of their factors. For example, the factorisation of x2 + 2x is x(x + 2), where x and x+2 are the factors that can be multiplied together to get the original polynomial.
Now let’s solve some factorisation problems here to understand better. Also, get some practice questions based on the factorisation at the end of the article.
Factorisation Problems and Solutions
Q.1: Factorise 4x2 + 12x + 5.
Solution: 4x2 + 12x + 5
We can write the above expression as:
⇒ 4x2 + 10x + 2x + 5
Taking the common terms out;
⇒ 2x(2x + 5) + 1 (2x + 5)
⇒ (2x + 1) (2x + 5)
Q.2: Factorise y2 + 16y + 60.
Solution: y2 + 16y + 60
We can write the above expression as:
⇒ y2 + 10y + 6y + 60
Taking the common terms out;
⇒ y(y + 10) + 6 (y + 10)
⇒ (y + 6) (y + 10)
Q.3: Factorise 5x2 + 14x – 3.
Solution: 5x2 + 14x – 3
We can write the above expression as:
⇒ 5x2 – x + 15x – 3
Taking the common terms out;
⇒ x(5x – 1) + 3(5x – 1)
⇒ (5x – 1) (x + 3)
Q.4: Factorise 4(x+y)2 – 28(x2 – y2) + 49(x-y)2.
Solution: Given,
4(x+y)2 – 28(x2 – y2) + 49(x-y)2
Since, by the algebraic formula we know;
x2 – y2 = (x + y) (x – y)
Therefore,
⇒ 4(x+y)2 – 28(x – y) (x + y) + 49(x-y)2
⇒ [2(x+y) – 7(x-y)]2
⇒ [2x + 2y – 7x – 7y]2
⇒ (9y – 5x)2
Q.5: Factorise 25(a+b)2 – (a-b)2.
Solution: 25(a+b)2 – (a-b)2
We can write the above expression as;
[5(a+b)]2 – (a-b)2By the algebraic formula, we know that, a2 – b2 = (a+b) (a-b).
Thus, we can write the expression as:
⇒[5(a+b)-(a-b)] [5(a+b)+(a-b)]
⇒ [5a + 5b – a + b] [5a + 5b + a – b]
⇒ [4a +6b] [6a + 4b]
Q.6: Factor 6a2b − 8ab + 10ab2.
Solution: In the given expression, the highest common factor is 2ab, therefore taking 2ab as common, we get;
6a2b − 8ab + 10ab2 = 2ab(3a − 4 + 5b)
Q.7: Solve (4x2 – 25y2)/(2x + 5y).
Solution: Given, (4x2 – 25y2)/(2x + 5y)
We can write, 4x2 – 25y2 as:
⇒ (2x)2 – (5y)2 = (2x + 5y) (2x – 5y)
Therefore,
⇒ [(2x + 5y) (2x – 5y)]/(2x + 5y)
⇒ 2x – 5y
Q.8: Factorise 2axy2 + 10x + 3ay2 + 15.
Solution: 2axy2 + 10x + 3ay2 + 15
Rearranging the terms we have;
⇒ (2axy2 + 3ay2) + (10x + 15)
Taking the common terms we get;
⇒ ay2(2x + 3) +5(2x + 3)
Hence, the required factors are:
⇒ (2x + 3) (ay2 + 5)
Q.9: Factorise x4 – (x – y)4.
Solution: Given, x4 – (x – y)4
It can be written as:
⇒ (x2)2 – [(x – y)2]2
Using algebraic formula, a2 – b2 = (a – b) (a + b)
⇒ [x2 – (x – y)2] [x2 + (x – y)2]
Using (a-b)2 = a2 – 2ab + b2, identity we get;
⇒ [x + (x – y] [x – (x – y)] [x2 + x2 – 2xy + y2]
⇒ (x + x – y) (x – x + y)[2x2 – 2xy + y2]
⇒ (2x – y) y(2x2 – 2xy + y2)
⇒ y(2x – y) (2x2 – 2xy + y2)
Q.10: Simplify the following:
Solution: Given,
We can write the above expression as:
Cancelling the common terms from numerator and denominator, we get;
= x – 2
Extra Questions for Factorisation
- Factor 18x3 +3x2 – 6x.
- Factorise 81x2 − 36 x + 4
- Find the factors of 7x2yz − 8y.
- Factor x2 − 11x + 24
- Factorise 11x2 + 33x − 110
- Factorise −16 + 49x2
- What are the factors of the expression 72 – 2x2?
- Find the factors of 9x2+4y2+12xy.
- Using the algebraic identities, factorise x2 + 6x + 9.
- If one of the factors of (5x2 + 70x – 160) is (x – 2), find the other factor.
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