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