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