The factorized form of $a^{2} + a c + b c - b^{2}$ is

(a - b) (a + b + c)

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(a + b) (a - b - c)

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(a + b) (a - b + c)

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(a - b) (a - b - c)

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Solution

The correct option is C $(a + b) (a - b + c)$
$a^{2} + a c + b c - b^{2}$
$= a^{2} - b^{2} + c (a + b)$
$= (a + b) (a - b + c)$

Suggest Corrections

Similar questions

\frac{a^{2} - b^{2} - c^{2}}{(a - b) (a - c)} + \frac{b^{2} - c^{2} - a^{2}}{(b - c) (b - a)} + \frac{c^{2} - a^{2} - b^{2}}{(c - a) (c - b)}

$a^{2} - b^{2}$ is equal to
$(a) (a - b)^{2}$
$(b) (a - b) (a - b)$
$(c) (a - b) (a + b)$
$(d) (a + b) (a + b)$

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

= (a - b) {(a - b) + 2 c} = (a - b) (a - b + 2 c)

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

a > 0; b > 0; c > 0, c - b \neq 1, c + b \neq 1,

{log}_{c + b} a + {log}_{c - b} a

\frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{b a}

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