Factor polynomial $$ 3x^{5}-192x^{3} $$
Answer
$$ 3x^{5}-192x^{3} = 3x^{3}( x-8 )( x+8 ) $$
Explanation
Step 1 :
After factoring out $ 3x^{3} $ we have:
$$ 3x^{5}-192x^{3} = 3x^{3} ( x^{2}-64 ) $$Step 2 :
Rewrite $ x^{2}-64 $ as:
$$ x^{2}-64 = (x)^2 - (8)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = x $ and $ II = 8 $ , we have:
$$ x^{2}-64 = (x)^2 - (8)^2 = ( x-8 ) ( x+8 ) $$This page was created using
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