Factorization Class 8 MCQ Questions

Answer: (a) (15a + 17b) × (15a – 17b)

Explanation: ​In order to factorize the difference of squares, we can use the formula\

a²– b²

=a² +ab – ab – b² *Adding and subtracting ab to the given equation

If we take a as common in ( a ² + ab) and-b as common in (-ab- b² ), we get

= (a+ b) × (a-b)

So, in the given expression we can write

(15a)² – (17b)² as,

= (15a+17b) x (15a-17b)

Result = (15a+17b) x (15a-17b)

Answer: (b) (5pq + 4rs) x (5pq – 4rs)

Explanation: ​In order to factorize the difference of squares, we can use the formula

a² – b²

= a² + ab – ab – b² *Adding and subtracting ab to the given equation

If we take a as common in (a² + ab) and-b as common in (-ab-b²), we get

= (a+ b) x (a-b)

So, in the given expression we can write

(5pq)² – (4rs)² as,

= (5pq+4rs) x (5pq-4rs)

Result = (5pq+4rs) x (5pq-4rs)

Answer: (d) (2m – 7)(2m – 5)

Explanation: ​For Quadratic trinomials we will first check the factors for constants of 1st and 3rd elements where their powers of variable (q) are 2 and 0.

Here, the constant of 1st term (m²) is 4 and the constant of last term (m⁰=1) is 35.

The product of the constants of 1st and

3rd elements is 4 and 35 is 4 X 35 = 140

We will factorize the product of the constants of 1st and 3rd term

= 2 x 2 x 5 x 7

If the third term is positive we would need to find two numbers, whose sum is the constant of 2nd term (i.e 24).

In this case 10 and 14 add up to 24

So, we can write the second term, 24m as 14m + 10m.

Accordingly, the equation would now become

4m² – (14m + 10m) + 35

4m² – 14m – 10m + 35

From 1st two term after expanding 2m is common and the last two term 5 is common. If we take 2m as common in (4m² – 14m) and 5 as common in (10m – 35), we get

= 2m x (2m – 7) – 5 x (2m – 7)

Since, 2m – 7 is common in both term, we can rewrite the equation as

= (2m – 5) x (2m – 7)

Hence the result = (2m – 7) (2m – 5)

Answer: (d) (2b-5)(6ab – 15a – 2)

Explanation: ​Firstly, we will observe the common binomials term

3a(2b−5)² – 2(2b-5)

Here, we observe that the binomial (2b – 5) is common to both the terms

Since, 2b – 5 is common in both term, we can rewrite the equation as

= (2b – 5) x {3a(2b – 5) – 2}

As, on of the term {3a(2b – 5) – 2} need to be simplified

= (2b – 5) x (6ab – 15a – 2)

Hence the result = (2b – 5) (6ab – 15a – 2)

Answer: (a) (12p² + 13q)²

Explanation: ​When the given equation is a perfect square, we have to arrange the term in a manner that they correspond to A² +B²± 2AB.

For perfect square we have to arrange the term in A² +B² ± 2AB

We know that, (A+B)²= A² +B² + 2AB

And, (A−B)² = A² +B² – 2AB

The given case, can also be written as,

144(p)⁴+169(q)² + 312p²q

= (12p²)²+ (13q)²+ 2 x 12(p)² x 13q

Comparing with the standard equation (A+B)²=

A² +B² + 2AB, we can say

A=12(p)²

B=13q

= (12p²+13q)²

Answer: (c) 14abc

Explanation: ​In order to find the HCF, we will first need to factorize the two terms

The first term can be written as 2 x 2 x 7 x a x a x a x b x b x c

The second term can be written as 2 x 7 x a x b x c x c

From the factors of the terms, we will need to find the common elements

Here, 2 x 7 x a x b x c are the common elements.

So, HCF is (2 x 7 x a x b x c) or 14abc