Factor polynomial $$ y^{2}-8y+16 $$
Answer
$$ y^{2}-8y+16 = ( y-4 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ x^2 = \left( \color{blue}{ y } \right)^2 ~~ \text{and} ~~ 16 = \left( \color{red}{ 4 } \right)^2 $$The middle term ( $ -8x $ ) is two times the product of the terms that are squared.
$$ -8x = - 2 \cdot \color{blue}{y} \cdot \color{red}{4} $$We can conclude that the polynomial $ y^{2}-8y+16 $ 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 = y } $ and $ \color{red}{ B = 4 } $ so,
$$ y^{2}-8y+16 = ( \color{blue}{ y } - \color{red}{ 4 } )^2 $$This page was created using
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