Factor polynomial $$ 3y^{2}-48 $$
Answer
$$ 3y^{2}-48 = 3( y-4 )( y+4 ) $$
Explanation
Step 1 :
After factoring out $ 3 $ we have:
$$ 3y^{2}-48 = 3 ( y^{2}-16 ) $$Step 2 :
Rewrite $ y^{2}-16 $ as:
$$ y^{2}-16 = (y)^2 - (4)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = y $ and $ II = 4 $ , we have:
$$ y^{2}-16 = (y)^2 - (4)^2 = ( y-4 ) ( y+4 ) $$This page was created using
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