Factorise:
$6 + 7 b - 3 b^{2}$

(2 - 3 b) (3 - b)

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(2 + 3 b) (3 - b)

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(2 + 3 b) (2 - b)

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(2 + 3 b) (2 + b)

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Solution

The correct option is C $(2 + 3 b) (3 - b)$
$F a c t o r i z i n g {- 3 b}^{2} + 7 b + 6, S p l i t 7 b a s : = - 3 b^{2} + 9 b - 2 b + 6 = 3 b (3 - b) + 2 (3 - b) = (3 - b) (3 b + 2)$

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Factorise :a²-(2a+3b)²

8 (2 a + 3 b)^{3} - 12 (2 a + 3 b)^{2}

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6 + 7 b - 3 b^{2}

[\frac{3}{4} a^{2} + 3 b^{2}] \times [4 (a^{2} - \frac{2}{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)

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