Factor polynomial $$ 121y^{2}-132y+36 $$
Answer
$$ 121y^{2}-132y+36 = ( 11y-6 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ 121x^2 = \left( \color{blue}{ 11y } \right)^2 ~~ \text{and} ~~ 36 = \left( \color{red}{ 6 } \right)^2 $$The middle term ( $ -132x $ ) is two times the product of the terms that are squared.
$$ -132x = - 2 \cdot \color{blue}{11y} \cdot \color{red}{6} $$We can conclude that the polynomial $ 121y^{2}-132y+36 $ 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 = 11y } $ and $ \color{red}{ B = 6 } $ so,
$$ 121y^{2}-132y+36 = ( \color{blue}{ 11y } - \color{red}{ 6 } )^2 $$This page was created using
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