Factor polynomial $$ 48q^{2}-72q+27 $$

$$ 48q^{2}-72q+27 = 3( 4q-3 )^{ 2 } $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 48q^{2}-72q+27 = 3 ( 16q^{2}-24q+9 ) $$

Step 2 :

Both the first and third terms are perfect squares.

$$ 16x^2 = \left( \color{blue}{ 4q } \right)^2 ~~ \text{and} ~~ 9 = \left( \color{red}{ 3 } \right)^2 $$

The middle term ( $ -24x $ ) is two times the product of the terms that are squared.

$$ -24x = - 2 \cdot \color{blue}{4q} \cdot \color{red}{3} $$

We can conclude that the polynomial $ 16q^{2}-24q+9 $ 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 = 4q } $ and $ \color{red}{ B = 3 } $ so,

$$ 16q^{2}-24q+9 = ( \color{blue}{ 4q } - \color{red}{ 3 } )^2 $$

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