# How do you find the product of (2n-7)^2?

Jun 29, 2016

$= {\left(2 n - 7\right)}^{2} = 4 {n}^{2} - 28 n + 49.$
Reqd. Product$= {\left(2 n - 7\right)}^{2} = \left(2 n - 7\right) \cdot \left(2 n - 7\right) = 2 n \left(2 n - 7\right) - 7 \left(2 n - 7\right) \ldots \left[D i s t r i b u t i v e L a w\right] = 4 {n}^{2} - 14 n - 14 n + 49 = 4 {n}^{2} - 28 n + 49.$
OR, we can use the formula, ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2} ,$ & get,
The product$= {\left(2 n\right)}^{2} - 2 \cdot 2 n \cdot 7 + {7}^{2} = 4 {n}^{2} - 28 n + 49 ,$ as before!