If $x^{2} - x (a + 2 b) + 2 a b$ is factorised, then its factors are

(x + a), (x + 2 b)

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(x + a), (x - 2 b)

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(x - a), (x + 2 b)

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(x - a), (x - 2 b)

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Solution

The correct option is D $(x - a), (x - 2 b)$
$x^{2} - x (a + 2 b) + 2 a b$
= $x^{2} - x a - 2 x b + 2 a b$
= $x (x - a) - 2 b (x - a)$
= $(x - a) (x - 2 b)$
$x = 2 b, x = a$

Suggest Corrections

Similar questions

If $(a - 2 b) - (2 b - a)$ = $x$ , then $x$ is

f (x) = a log | x | + b x^{2} + x

x = - 1

x = 2

a

b

x = \frac{4 a b}{a + b}

\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b}

t a n x = \frac{2 b}{a - c} (a \neq c), y = a c o s^{2} x + 2 b s i n x c o s x + c s i n^{2} x a n d z = a s i n^{2} x - 2 b s i n x c o s x + c o s^{2} x,

x = \frac{4 a b}{a + b}

\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b} = 2

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