SOLUTION: 81x^4-16y^4 (p+8q)^2-10(p+8q)+25

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Question 68560: 81x^4-16y^4
(p+8q)^2-10(p+8q)+25
Found 2 solutions by funmath, stanbon:
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
81x^4-16y^4
These are easier if you remember that to factor the difference of perfect squares:
highlight%28a%5E2-b%5E2=%28a%2Bb%29%28a-b%29%29
%289x%5E2%29%5E2-%284y%5E2%29%5E2 a=9x^2 and b=4y^2
%289x%5E2%2B4y%5E2%29%289x%5E2-4y%5E2%29
The sum of perfect squares is prime, but the second set of parentheses is the difference of perfect squares and can be factored:
%289x%5E2%2B4y%5E2%29%28%283x%29%5E2-%282y%29%5E2%29 a=3x and b=2y
highlight%28%289x%5E2%2B4y%5E2%29%283x%2B2y%29%283x-2y%29%29
:
(p+8q)^2-10(p+8q)+25
These are easier if you remember that a perfect square trinomial can be factored:
highlight%28a%5E2-2ab%2Bb%5E2=%28a-b%29%5E2%29
%28p%2B8q%29%5E2-2%2A%28p%2B8q%29%2A5%2B5%5E2 a=(p+8q) and b=5
%28%28p%2B8q%29-5%29%5E2
highlight%28%28p%2B8q-5%29%5E2%29
:
Happy Calculating!!!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
81x^4-16y^4
Rewrite as: (3x)^4 -(2y)^4
Factor as: [(3x)^2 + (2y)^2][(3y+2y)][(3x)-(2x)]
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Cheers,
Stan H.