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