Factor polynomial $$ 3y^{2}+24y+48 $$

$$ 3y^{2}+24y+48 = 3( y+4 )^{ 2 } $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 3y^{2}+24y+48 = 3 ( y^{2}+8y+16 ) $$

Step 2 :

Both the first and third terms are perfect squares.

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

The middle term ( $ 8x $ ) is two times the product of the terms that are squared.

$$ 8x = 2 \cdot \color{blue}{y} \cdot \color{red}{4} $$

We can conclude that the polynomial $ y^{2}+8y+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 = y } $ and $ \color{red}{ B = 4 } $ so,

$$ y^{2}+8y+16 = ( \color{blue}{ y } + \color{red}{ 4 } )^2 $$

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