Factorize the following equation by the grouping method
$a^{2} + 4 b^{2} - 3 a + 6 b - 4 a b$

(a - b) (a - b - b c)

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(a - 2 b) (a - 2 b - 3)

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(a - b) (a - 2 b - b c)

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(a - 2 b) (a - b - 3)

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Solution

The correct option is A $(a - 2 b) (a - 2 b - 3)$
$a^{2} + 4 b^{2} - 3 a + 6 b - 4 a b$
$= a^{2} - 4 a b + 4 b^{2} - 3 a + 6 b$
$= (a - 2 b)^{2} - 3 (a - 2 b)$
$= (a - 2 b) (a - 2 b - 3)$

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∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ =

∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + 2 b a + 2 b & a & a + b a + b & a + 2 b & a \end{matrix} ∣ ∣ ∣ ∣ = 9 b^{2} (a + b)

a^{2} + 4 b^{2} - 3 a + 6 b - 4 a b

∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣

\frac{a}{a + 2 b} = \frac{a - 2 b}{a - 4 c}

\frac{b}{b + c} = \frac{a - b}{a - c}

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