Factor polynomial $$ 64n^{2}+208n+169 $$
Answer
$$ 64n^{2}+208n+169 = ( 8n+13 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ 64x^2 = \left( \color{blue}{ 8n } \right)^2 ~~ \text{and} ~~ 169 = \left( \color{red}{ 13 } \right)^2 $$The middle term ( $ 208x $ ) is two times the product of the terms that are squared.
$$ 208x = 2 \cdot \color{blue}{8n} \cdot \color{red}{13} $$We can conclude that the polynomial $ 64n^{2}+208n+169 $ 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 = 8n } $ and $ \color{red}{ B = 13 } $ so,
$$ 64n^{2}+208n+169 = ( \color{blue}{ 8n } + \color{red}{ 13 } )^2 $$This page was created using
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