Factorise a-4 - b-4 - Maths Q-A

Factorise the polynomiala–b–a^2+b^20ex0ex=a b 1(a^2 b^2)using the identity x^2 y^2=(x+y)(x y)we get=a b (a b)(a+b)0ex0ex=(a b)[1 (a+b)]Hence, the factor of a–b– ...

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Standard IX

Mathematics

Factorisation Using Algebraic Identities

Factorise a^4...

Question

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

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Solution

Factorise the polynomial
$a - b - a^{2} + b^{2} = a - b - 1 (a^{2} - b^{2})$
using the identity $x^{2} - y^{2} = (x + y) (x - y)$ we get
$= a - b - (a - b) (a + b) = (a - b) [1 - (a + b)]$
Hence, the factor of $a - b - a^{2} + b^{2}$ are $(a - b)$ and $[1 - (a + b)]$ .

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Factorise: a–b–a2+b2

Factorise the polynomiala–b–a2+b2=a-b-1(a2-b2)using the identity x2-y2=(x+y)(x-y)we get=a-b-(a-b)(a+b)=(a-b)1-(a+b)Hence, the factor of a–b–a2+b2 are (a-b)and 1-(a+b).

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

Factorise the polynomial
$a - b - a^{2} + b^{2} = a - b - 1 (a^{2} - b^{2})$
using the identity $x^{2} - y^{2} = (x + y) (x - y)$ we get
$= a - b - (a - b) (a + b) = (a - b) [1 - (a + b)]$
Hence, the factor of $a - b - a^{2} + b^{2}$ are $(a - b)$ and $[1 - (a + b)]$ .