Factor polynomial $$ -162m^{5}+32m $$
Answer
$$ -162m^{5}+32m = 2m( -3m+2 )( 3m+2 )( 9m^{2}+4 ) $$
Explanation
Step 1 :
After factoring out $ 2m $ we have:
$$ -162m^{5}+32m = 2m ( -81m^{4}+16 ) $$Step 2 :
Rewrite $ -81m^{4}+16 $ as:
$$ -81m^{4}+16 = 16 -81x^4 = (4)^2 - (9m^{2})^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4 $ and $ II = 9m^{2} $ , we have:
$$ -81m^{4}+16 = (4)^2 - (9m^{2})^2 = ( -9m^{2}+4 ) ( 9m^{2}+4 ) $$Step 3 :
Rewrite $ -9m^{2}+4 $ as:
$$ -9m^{2}+4 = 4 -9x^2 = (2)^2 - (3m)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2 $ and $ II = 3m $ , we have:
$$ -9m^{2}+4 = (2)^2 - (3m)^2 = ( -3m+2 ) ( 3m+2 ) $$This page was created using
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