Question
Factor polynomial $$ 16x^{8}y^{4}-81z^{4} $$
Answer
$$ 16x^{8}y^{4}-81z^{4} = (4x^{4}y^{2}+9z^{2})(2x^{2}y+3z)(2x^{2}y-3z) $$
Explanation
Step 1 :
Rewrite $ 16x^8y^4-81z^4 $ as:
$$ \color{blue}{ 16x^8y^4-81z^4 = (4x^4y^2)^2 - (9z^2)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4x^4y^2 $ and $ II = 9z^2 $ , we have:
$$ 16x^8y^4-81z^4 = (4x^4y^2)^2 - (9z^2)^2 = ( 4x^4y^2-9z^2 ) ( 4x^4y^2+9z^2 ) $$Step 2 :
Rewrite $ 4x^4y^2-9z^2 $ as:
$$ \color{blue}{ 4x^4y^2-9z^2 = (2x^2y)^2 - (3z)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2x^2y $ and $ II = 3z $ , we have:
$$ 4x^4y^2-9z^2 = (2x^2y)^2 - (3z)^2 = ( 2x^2y-3z ) ( 2x^2y+3z ) $$This page was created using
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