Factor polynomial $$ 81x^{2}-72x+16 $$
Answer
$$ 81x^{2}-72x+16 = ( 9x-4 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ 81x^2 = \left( \color{blue}{ 9x } \right)^2 ~~ \text{and} ~~ 16 = \left( \color{red}{ 4 } \right)^2 $$The middle term ( $ -72x $ ) is two times the product of the terms that are squared.
$$ -72x = - 2 \cdot \color{blue}{9x} \cdot \color{red}{4} $$We can conclude that the polynomial $ 81x^{2}-72x+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 = 9x } $ and $ \color{red}{ B = 4 } $ so,
$$ 81x^{2}-72x+16 = ( \color{blue}{ 9x } - \color{red}{ 4 } )^2 $$This page was created using
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