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Prime Factorisation

Factorise:2 a...

Question

Factorise:
(2a + 3b)² − 16c²

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Solution

We have:
${(2 a + 3 b)}^{2} - 16 c^{2} = {(2 a + 3 b)}^{2} - {(4 c)}^{2} = \{(2 a + 3 b) + 4 c\} \{(2 a + 3 b) - 4 c\} = (2 a + 3 b + 4 c) (2 a + 3 b - 4 c)$

∴ ${(2 a + 3 b)}^{2} - 16 c^{2} = (2 a + 3 b + 4 c) (2 a + 3 b - 4 c)$

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Similar questions

Q. Factorise:
2a + 6b − 3(a + 3b)²

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

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

121 b^{2} - 88 b c + 16 c^{2}

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