Question
The factors of (a2 + 4b2 + 4b - 4ab - 2a - 8) are?
Answer: Option A
$$\eqalign{
& {a^2} + 4{b^2} + 4b - 4ab - 2a - 8 \cr
& = {a^2} - 4ab + 4{b^2} - 2a + 4b - 8 \cr
& = {\left( {a - 2b} \right)^2} - 2\left( {a - 2b} \right) - 8 \cr
& {\text{Put }} t = a - 2b \cr
& = {t^2} - 2t - 8 \cr
& = {t^2} - 4t + 2t - 8 \cr
& = t\left( {t - 4} \right) + 2\left( {t - 4} \right) \cr
& = \left( {t + 2} \right)\left( {t - 4} \right) \cr
& = \left( {a - 2b - 4} \right)\left( {a - 2b + 2} \right) \cr
& \left( {{\text{Put the value of assume }}t} \right) \cr} $$
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$$\eqalign{
& {a^2} + 4{b^2} + 4b - 4ab - 2a - 8 \cr
& = {a^2} - 4ab + 4{b^2} - 2a + 4b - 8 \cr
& = {\left( {a - 2b} \right)^2} - 2\left( {a - 2b} \right) - 8 \cr
& {\text{Put }} t = a - 2b \cr
& = {t^2} - 2t - 8 \cr
& = {t^2} - 4t + 2t - 8 \cr
& = t\left( {t - 4} \right) + 2\left( {t - 4} \right) \cr
& = \left( {t + 2} \right)\left( {t - 4} \right) \cr
& = \left( {a - 2b - 4} \right)\left( {a - 2b + 2} \right) \cr
& \left( {{\text{Put the value of assume }}t} \right) \cr} $$
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