Factorisation| Worksheet Solution| Class 8

Worksheet solutions for chapter: Factorisation for class 8 are provided below.

(1) Factorise -16a2 + 2a – 8a2 + a

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(2)Factorise the following expression: 4a2 + 16a + 16

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Find the factors of given expressions.

(3)81x2 – y2

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(4) x2 – 14x + 49

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x2 – 14x + 49

In above expression

a = x, b = 7, 2ab = 2 × 7 × b

Put values in identity a2 – 2ab + b2 =(a – b)2

x2 – 14x + 49 = (x – 7)2 Required factorization

(5) 8x2 – 2y2

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(6) x2 + 8x + 15

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x2 + 8x + 15

In the above expression

x2 = x2, ab = 15 and (a+b) = 8

ab = 15, a and b are factors of 15.

Factors of 15 = 3 × 5

Sum of 3 and 5 = 3 + 5 = 8

Therefore, a and b should be  3 and 5.

Now put the values in the identity (x + a ) (x + b) = x2 + (a + b)x + ab

x2 + (3 + 5)x + 15 = x2 + 3x + 5x + 15

Taking common factor out

x2 + (3 + 5)x + 15 = x(x + 3) + 5(x +3)

Taking (x + 3) out from the terms

x2 + (3 + 5)x + 15 = (x + 3) ( x + 5) Required factorisation

(7) Divide (x2 – 6x -7) ÷ (x – 1)

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(8)Find error and correct it in given expression.

(5a + b)2 = 5a2 + 10ab + b2

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(5a + b)2 = 5a2 + 10ab + b2

(5a + b)2 = (5a)2 + 2 (5a) (b) + b2[ (a + b)2 = a2 + 2ab + b2]

(5a + b)2 = 25a2 + 10ab + b2

Therefore, error: Identity (a + b)2 = a2 + 2ab + b2 not used properly.

Correction: square of 5a is 25a2

Hence, correct expression is (5a + b)2 = 25a2 + 10ab + b2

(9)Find the common factor for given expression and choose correct answer.

7a2b4 + 14b2c4 + 21c2a4

(a) 7a2 b2

(b) a2b4

(c)7

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Factorisation, Q 10, practice sheet, NCERT, class 8

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Helping Topics

Factorisation

NCERT Solutions Class 8