Factor polynomial $$ b^{2}+16b+64 $$
Answer
$$ b^{2}+16b+64 = ( b+8 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ x^2 = \left( \color{blue}{ b } \right)^2 ~~ \text{and} ~~ 64 = \left( \color{red}{ 8 } \right)^2 $$The middle term ( $ 16x $ ) is two times the product of the terms that are squared.
$$ 16x = 2 \cdot \color{blue}{b} \cdot \color{red}{8} $$We can conclude that the polynomial $ b^{2}+16b+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 = b } $ and $ \color{red}{ B = 8 } $ so,
$$ b^{2}+16b+64 = ( \color{blue}{ b } + \color{red}{ 8 } )^2 $$This page was created using
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