Factor polynomial $$ t^{2}-18t+81 $$
Answer
$$ t^{2}-18t+81 = ( t-9 )^{ 2 } $$
Explanation
Both the first and third terms are perfect squares.
$$ x^2 = \left( \color{blue}{ t } \right)^2 ~~ \text{and} ~~ 81 = \left( \color{red}{ 9 } \right)^2 $$The middle term ( $ -18x $ ) is two times the product of the terms that are squared.
$$ -18x = - 2 \cdot \color{blue}{t} \cdot \color{red}{9} $$We can conclude that the polynomial $ t^{2}-18t+81 $ 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 = t } $ and $ \color{red}{ B = 9 } $ so,
$$ t^{2}-18t+81 = ( \color{blue}{ t } - \color{red}{ 9 } )^2 $$This page was created using
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