A factor nbsp-of a-2-2ab-b-2-c-2 is -a-b-c-a-b-c-Both 1 and 2-a-b-c-

The correct option is C Both 1 and 2 a^2 2ab+b^2 c^2 = (a b)^2 – c^2 Using identity x^2 y^2 = nbsp; (x+y)(x y) nbsp; we get (a b)^2 – c^2 = [(a b)+c] [( ...

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Standard VIII

Mathematics

Factorisation using algebraic identities

A factor nb...

Question

A factor of $a^{2} - 2 a b + b^{2} - c^{2}$ is ___________.

a-b+c

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a-b-c

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Both 1 and 2

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a+b-c

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Solution

The correct option is C
Both 1 and 2

$a^{2} - 2 a b + b^{2} - c^{2} = (a - b)^{2} - c^{2}$
Using identity $x^{2} - y^{2} = (x + y) (x - y)$ we get

$(a - b)^{2} - c^{2}$ = $[(a - b) + c] [(a - b) - c]$

Hence, factors of $a^{2} - 2 a b + b^{2} - c^{2}$ are $a - b + c$ and $a - b - c$

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Similar questions

Divide:
$[(a^{2} + 2 a b + b^{2}) - (a^{2} + 2 a c + c^{2})]$ by $2 a + b + c$

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

Factorise:

$(a - b + c)^{2} + (b - c + a)^{2} + 2 (a - b + c) (a - c + a)$

Q. If

a + b + c = m

and

a^{2} + b^{2} + c^{2} = n

,
then

2 (a b + b c + c a)

is _______.

Q. Divide:
(a² + 2ab + b²) − (a² + 2ac + c²) by 2a + b + c

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A factor of a2−2ab+b2−c2 is ___________.

The correct option is C Both 1 and 2 a2−2ab+b2−c2=(a−b)2–c2 Using identity x2−y2=(x+y)(x−y) we get (a−b)2–c2 = [(a−b)+c][(a−b)−c]​ Hence, factors of a2−2ab+b2−c2 are a−b+c and a−b−c

A factor of $a^{2} - 2 a b + b^{2} - c^{2}$ is ___________.

The correct option is C
Both 1 and 2

$a^{2} - 2 a b + b^{2} - c^{2} = (a - b)^{2} - c^{2}$
Using identity $x^{2} - y^{2} = (x + y) (x - y)$ we get

$(a - b)^{2} - c^{2}$ = $[(a - b) + c] [(a - b) - c]$

Hence, factors of $a^{2} - 2 a b + b^{2} - c^{2}$ are $a - b + c$ and $a - b - c$