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Factorisation by Regrouping Terms

Factorise : 2...

Question

Factorise :
$2 a + 6 b - 3 (a + 3 b)^{2}$

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Solution

$2 a + 6 b - 3 (a + 3 b)^{2} = 2 (a + 3 b) - 3 (a + 3 b)^{2} = (a + 3 b) {2 - 3 (a + 3 b)} = (a + 3 b) (2 - 3 a - 9 b) A n s .$

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2 a + 6 b - 3 (a + 3 b)^{2}

2 a + 6 b - 3 (a + 3 b)^{2}

= 2 (a + 3 b) - 3 (a + 3 b)^{2} [T a k i n g 2 c o m m o n f r o m 2 a + 6 b]

= (a + 3 b) {\begin{matrix} 2 - 3 (a + 3 b) \end{matrix}} [T a k i n g (a + 3 b) c o m m o n]

= (a + 3 b) (2 - 3 a - 9 b)

2 a + 6 b - 3 (a + 3 b)^{2}

Q. Factorise:
(2a + 3b)² − 16c²

a^{2} - (2 a + 3 b)^{2}

(2 a + 3 b)^{2} - 5 (2 a + 3 b) - 14

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Factorisation by Regrouping Terms

Standard VIII Mathematics

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