Question

Factorize the following expression: $121b^2-88bc+16c^2$

Answer 1

• The expression $121b^2-88bc+16c^2$ can be written as ${\left(11b\right)}^2-2\left(11b\right)\left(4c\right)+{\left(4c\right)}^2$ ….[1]
• We know the Algebraic identity ${\left(a-b\right)}^2=a^2-2ab+b^2$ ….[2]
• By comparing [1] and [2], we can say, $121b^2-88bc+16c^2$$=$${\left(11b\right)}^2-2\left(11b\right)\left(4c\right)+{\left(4c\right)}^2$

$={\left(11b-4c\right)}^2$

Hence, $\mathbf{121}{\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{88}\mathit{b}\mathit{c}\mathbf{+}\mathbf{16}{\mathit{c}}^{\mathbf{2}}\mathbf{=}\left(11b-4c\right)\mathbf{(}\mathbf{11}\mathbf{b}\mathbf{-}\mathbf{4}\mathbf{c}\mathbf{)}$