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Standard X

Mathematics

Factorisation of Quadratic Polynomials

Factorise: ...

Question

Factorise: $a - b - a^{2} + b^{2}$

(a - b) [1 - (a + b)]

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(a + b) [1 - (a - b)]

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(a - b) [1 - (a - b)]

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none of these

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Solution

The correct option is A $(a - b) [1 - (a + b)]$
Given polynomial is
$a - b - a^{2} + b^{2} = (a - b) - (a^{2} - b^{2})$
$= (a - b) - (a - b) (a + b)$
$= (a - b) [1 - (a + b)]$
Option A is correct.

Suggest Corrections

Similar questions

Factorize:

$a^{2} - b^{2} - 2 b - 1$

Q. Factorise:

a^{2} - b^{2} - a + b

If $\frac{1}{b - c}$ , $\frac{1}{c - a}$ , $\frac{1}{a - b}$ be consecutive terms of an A.P., then $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ will be in ___.

Q. Which step in the following problem is wrong?

a = b = 1

a = b

step-1:

a^{2} = a b

step-2:

a^{2} - b^{2} = a b - b^{2}

step-3:

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

step-4: a + b =

\frac{b (a - b)}{a - b}

a + b = b

1 + 1 = 1

2 = 1

Q. If

A = [\begin{matrix} 1 & - 1 2 & - 1 \end{matrix}], B = [\begin{matrix} a & 1 b & - 1 \end{matrix}]

and

(A + B)^{2} = A^{2} + B^{2}

,find

a

and

b

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Factorise: a−b−a2+b2

The correct option is A (a−b)[1−(a+b)]Given polynomial is a−b−a2+b2=(a−b)−(a2−b2) =(a−b)−(a−b)(a+b) =(a−b)[1−(a+b)] Option A is correct.

Factorise: $a - b - a^{2} + b^{2}$

The correct option is A $(a - b) [1 - (a + b)]$
Given polynomial is
$a - b - a^{2} + b^{2} = (a - b) - (a^{2} - b^{2})$
$= (a - b) - (a - b) (a + b)$
$= (a - b) [1 - (a + b)]$
Option A is correct.