NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation

Question 3 Exercise 14.2

Factorise the expression:

121 b^{2}-88 b c+16 c^{2}



The factorization is made much simpler by employing several identities.

The given expression needs to be factored and is of the form or may be written as: a2 - 2ab + b2 = (a-b)2.

\begin{array}{l} 121 \mathrm{b}^{2}-88 \mathrm{bc}+16 \mathrm{c}^{2} \\ =(11 b)^{2}-2 \times 11 b \times 4 c+(4 c)^{2} \\ =(11 b-4 c)^{2} \\ \text { Using identity: }(x-y)^{2}=x^{2}-2 x y+y^{2} \end{array}

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