SOLUTION: factor completely with respect to the intergers 10x^4- 160

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: factor completely with respect to the intergers 10x^4- 160      Log On


   

Question 59003: factor completely with respect to the intergers 10x^4- 160
Found 2 solutions by stanbon, funmath:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
10x^4- 160
=10(x^4-16)
=10(x^2+4)(x^2-4)
=10(x^2+4)(x+2)(x-2)
Cheers,
Stan H.

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Two rules to remember on this one:
highlight%28a%5E2-b%5E2=%28a%2Bb%29%28a-b%29%29 Factorization of difference of perfect squares
highlight%28a%5E2%2Bb%5E2=prime%29 Sums of perfect squares are prime.
:
factor completely with respect to the intergers
10x%5E4-+160 Factor out GCF of 10
10%28x%5E4-16%29 Factor the difference of perfect squares a=x^2 and b=4.
10%28x%5E2%2B4%29%28x%5E2-4%29 Factor the difference of perfect squares a=x and b=2.
10%28x%5E2%2B4%29%28x%2B2%29%28x-2%29
Happy Calculating!!!