Factorise the following by grouping the terms:

r) $2 p (a^{2} - 2 b^{2}) - 14 p + (a^{2} - 2 b^{2})^{2} - 7 (a^{2} - 2 b^{2})$

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Solution

Given:

$2 p (a^{2} - 2 b^{2}) - 14 p + (a^{2} - 2 b^{2})^{2} - 7 (a^{2} - 2 b^{2})$

$= 2 p (a^{2} - 2 b^{2}) + (a^{2} - 2 b^{2})^{2} - 14 p - 7 (a^{2} - 2 b^{2})$

On grouping the terms, we get,

$= (2 p (a^{2} - 2 b^{2}) + (a^{2} - 2 b^{2})^{2}) - (14 p + 7 (a^{2} - 2 b^{2}))$

$= (a^{2} - 2 b^{2}) (2 p + a^{2} - 2 b^{2}) - 7 (2 p + a^{2} - 2 b^{2})$

$= (2 p + a^{2} - 2 b^{2}) (a^{2} - 2 b^{2} - 7)$

Hence the answer is $(2 p + a^{2} - 2 b^{2}) (a^{2} - 2 b^{2} - 7)$

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\frac{7 a^{2} + 2 b^{2}}{7 a^{2} + 2 b^{2}} = \frac{113}{13}

\frac{a}{b}

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$8 (3 a - 2 b)^{2} - 10 (3 a - 2 b)$

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\frac{7 a^{2} + 2 b^{2}}{7 a^{2} - 2 b^{2}} = \frac{113}{13}

\frac{a}{b}

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