Factor polynomial $$ 81n^{2}+180n+100 $$
Answer
$$ 81n^{2}+180n+100 = ( 9n+10 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ 81x^2 = \left( \color{blue}{ 9n } \right)^2 ~~ \text{and} ~~ 100 = \left( \color{red}{ 10 } \right)^2 $$The middle term ( $ 180x $ ) is two times the product of the terms that are squared.
$$ 180x = 2 \cdot \color{blue}{9n} \cdot \color{red}{10} $$We can conclude that the polynomial $ 81n^{2}+180n+100 $ 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 = 9n } $ and $ \color{red}{ B = 10 } $ so,
$$ 81n^{2}+180n+100 = ( \color{blue}{ 9n } + \color{red}{ 10 } )^2 $$This page was created using
Polynomial Factoring Calculator