SOLUTION: #1)The problem is 3x^4- 81x, I could only factor out 3x to get 3x(x^3-27),is this as far as I can go? #2)3t^2+t=10,I put the problem to zero 3t^2+t-10=0,next I was stuck,is this

Question 156106: #1)The problem is 3x^4- 81x, I could only factor out 3x to get 3x(x^3-27),is this as far as I can go?
#2)3t^2+t=10,I put the problem to zero 3t^2+t-10=0,next I was stuck,is this unfactorable?
#3)Could you please check my work on this problem 4x^2y^2-4x^2-9y^2+9 I grouped and factored 4x^2(y^2-1)-9(y^2-1),(4x^2)(y^2-1) then I factored these out (2x+3)(2x-3)(y-1)(y+1)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
#1)The problem is 3x^4- 81x, I could only factor out 3x to get 3x(x^3-27),is this as far as I can go?
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x^3-27 is called "the difference of cubes"; it can be factored as follows:
(x-3)(x^2 + 3x + 9)
Final answer: 3x(x-3)(x^2 + 3x + 9)
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#2)3t^2+t=10,I put the problem to zero
3t^2+t-10=0,next I was stuck,is this unfactorable?
3t^2 +6t-5t -10 = 0
3t(t+2) -5(t+2) = 0
(t+2)(3t-5) = 0
t= -2 or t = 5/3
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#3)Could you please check my work on this problem
4x^2y^2-4x^2-9y^2+9 I grouped and factored 4x^2(y^2-1)-9(y^2-1),(4x^2-9)(y^2-1) then I factored these out (2x+3)(2x-3)(y-1)(y+1)
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Good.
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Cheers,
Stan H.

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