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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