Factorise:-
$2 y^{3} + y^{2} - 2 y - 1$ .

(y - 1) (y + 1) (2 y + 1)

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(y + 1) (y + 1) (2 y + 1)

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(y - 1) (y + 2) (2 y + 1)

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(y - 1) (y - 1) (2 y + 1)

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Solution

The correct option is D $(y - 1) (y + 1) (2 y + 1)$
$2 y^{3} + y^{2} - 2 y - 1$

$\Rightarrow y^{2} (2 y + 1) - 1 (2 y + 1)$

$\Rightarrow (y^{2} - 1) (2 y + 1)$

$\Rightarrow (y^{2} - 1^{2}) (2 y + 1)$

$[∵ 1^{2} = 1]$

$\Rightarrow (y + 1) (y - 1) (2 y + 1)$

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\frac{1}{2 x} - \frac{1}{y} + 1 = 0; \frac{1}{x} + \frac{1}{2 y} = 8, x \neq 0, y \neq 0

$y^{2} + 2 y - 3 = ? (a) (y - 1) (y + 3) (b) (y + 1) (y - 3) (c) (y - 1) (y - 3) (d) (y + 2) (y - 3)$

Using factor theorem the factors of $y^{3} - 2 y^{2} - y + 2$ is

\frac{y}{y - 2} + \frac{1}{y + 2} + \frac{3 y}{4 - y^{2}}

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