Important Questions Class 8 Maths Chapter 14 Factorization

Class 8 important questions for chapter 14 factorization provided here cover various short answer type questions, long answer type questions and HOTS questions. These factorization questions will help CBSE students to be well-prepared for class 8 exam. Here, some important questions related to factorization topic from NCERT class 8 book are also included.

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Factorization Important Questions For Class 8 (Chapter 14)

Some extremely important class 8 questions from chapter 14 i.e. factorization are provided below.

Short Answer Type Questions:

1. Express the following as in the form of (a+b)(a-b)

(i) a2 – 64

(ii) 20a2 – 45b2

(iii) 36x2y2 – 8

(iv) x2 – 2xy + y2 – z2

(v) 49x2 – 1

Solution:

For representing the expressions in (a+b)(a-b) form, use the following formula

a2 – b2 = (a+b)(a-b)

(i) a2 – 64 = a2 – 82 = (a + 9)(a – 8)

(ii) 20a2 – 45b2 = 5(4a2 – 9b2) = 5(2a + 3b)(2a – 3b)

(iii) 36x2y2 – 8 = 8( 4x2y2 – 1) = 8(2xy + 1)(2xy – 1)

(iv) x2 – 2xy + y2 – z2 = (x – y)2 – z2 = (x – y – z)(x – y + z)

(v) 49x2 – 1 = 72 – 12 = (7 + 1)(7 – 1)

2. Verify whether the following equations are correct. Rewrite the incorrect equations correctly.

(i) (a + 6)2 = a2 + 12a + 36

(ii) (2a)2 + 5a = 4a + 5a

Solution:

(i) (a + 6)2 = a2 + 12a + 36

Here, LHS = (a + 6)2 = a2 + 12a + 36

Now, RHS = a2 + 12a + 36

Hence, LHS = RHS.

(ii) (2a)2 + 5a = 4a + 5a

Here, LHS = (2a)2 + 5a = 4a2 + 5a

Now, RHS = 4a + 5a

So, LHS ≠ RHS

Correct equation: (2a)2 + 5a = 4a2 + 5a

3. For a = 3, simplify a2 + 5a + 4 and a2 – 5a

Solution:

Substitute the value of a = 3 in the given equations.

a2 + 5a + 4 = 32 + 5(3) + 4 = 9 + 15 + 4 = 28

And,

a2 – 5a = 32 – 5(3) = 9 – 15 = -6
 

Long Answer Type Questions:

4. Find the common factors of the following:

(i) 6 xyz, 24 xy2 and 12 x2y

(ii) 3x2 y3, 10x3 y2 and 6x2 y2 z

Solution:

(i) 6 xyz = 2 × 3 × x × y × z

24 xy2 = 2 × 2 × 2 × 3 × x × y × y

12 x2y = 2 × 2 × 3 × x × x × y

Thus, the common factors are common factors of 6 xyz, 24 xy2 and 12 x2y are 2, 3, x, y and, (2 × 3 × x × y) = 6xy

(ii) 3x2 y3 = 3 × x × x × y × y × y

10x3 y2 = 2 × 5 × x × x × x × y × y

6 x2 y2 z = 3 × 2 × x × x × y × y × z

Now, the common factors of 3x2 y3, 10x3 y2 and 6x2 y2 z are x2, y2 and, (x2 × y2) = x2 y2

5. Factroize the following expressions:

(i) 54x3y + 81x4y2

(ii) 14(3x – 5y)3 + 7(3x – 5y)2

(iii) 15xy + 15 + 9y + 25x

Solution:

(i) 54x3y + 81x4y2

= 2 × 3 × 3 × 3 × x × x × x × y + 3 × 3 × 3 × 3 × x × x × x × x × y × y

= 3 × 3 × 3 × x × x × x × y × (2 + 3 xy)

= 27x3y (2 + 3 xy)

(ii) 14(3x – 5y)3 + 7(3x – 5y)2

= 7(3x – 5y)2 [2(3x – 5y) +1]

= 7(3x – 5y)2 (6x – 10y + 1)

(iii) 15xy + 15 + 9y + 25x

Rearrange the terms as:

15xy + 25x + 9y + 15

= 5x(3y + 5) + 3(3y + 5)

Or, (5x + 3)(3y + 5)

6. Factorize (x + y)2 – 4xy

Solution:

To solve this expression, expand (x + y)2

Use the formula:

(𝑥 + 𝑦)2 = 𝑥2 + 2𝑥𝑦 + 𝑦2

(x + y)2 – 4xy = 𝑥2 + 2𝑥𝑦 + 𝑦2 – 4xy

= 𝑥2 + 𝑦2 – 2xy

We know, (x – y)2 = 𝑥2 + 𝑦2 – 2xy

So, factorization of (x + y)2 – 4xy = (x – y)2

7. Factorize x2 + 6x – 16

Solution:

To factorize, it should be checked that the sum of factors of 16 should be equal to 6.

Here, 16 = -2 × 8 and 8 + (-2) = 6

So,

x2 + 6x – 16 = x2 – 2x + 8x – 16

= x(x – 2) + 8(x – 2)

= (x + 8) (x – 2)

Hence, x2 + 6x – 16 = (x + 8) (x – 2)

8. Solve for (4x2 – 100) ÷ 6(x + 5)

Solution:

Important Questions Class 8 Maths Chapter 14 Factorization

= ⅔ (x – 5)

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