Q1 Factorise the following expressions

(VI)

121 b^2 - 88 bc + 16 c^2

Answers (1)

We have,
             121 b^2 - 88 bc + 16 c^2= 121b^2 - 44bc - 44bc + 16c^2
                                                       = 11b(11b - 4c) - 4c(11b - 4c)
                                                        = (11b-4c)(11b-4c) =(11b -4c)^{2}
Therefore,
                  121 b^2 - 88 bc + 16 c^2 = (11b -4c)^{2}

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