Factor polynomial $$ w^{2}-16w+64 $$
Answer
$$ w^{2}-16w+64 = ( w-8 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ x^2 = \left( \color{blue}{ w } \right)^2 ~~ \text{and} ~~ 64 = \left( \color{red}{ 8 } \right)^2 $$The middle term ( $ -16x $ ) is two times the product of the terms that are squared.
$$ -16x = - 2 \cdot \color{blue}{w} \cdot \color{red}{8} $$We can conclude that the polynomial $ w^{2}-16w+64 $ is a perfect square trinomial, so we will use the formula below.
$$ A^2 - 2AB + B^2 = (A - B)^2 $$In this example we have $ \color{blue}{ A = w } $ and $ \color{red}{ B = 8 } $ so,
$$ w^{2}-16w+64 = ( \color{blue}{ w } - \color{red}{ 8 } )^2 $$This page was created using
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