1 final question... and I am going to sleep.... factoring issues ... I am locked after this one section.... of the problem...
x^4+2x^3-3x-6 - problem
This is what I have so far and then I for some reason get stuck...
x(x^3+2x^3-3-6) (step 1)
x(X^3+2x^2-9) (step 2) - now I am stuck.... blaaaah!!!!
what am I missing because I surely can't figure it out...
2007-11-28 17:26:16 · 4 answers · asked by aunt24 1 in Science & Mathematics ➔ Mathematics
AHA!!!! thanks all for your help.... i see now where I mixed it around and messed it up... thanks again......
2007-11-28 17:36:11 · update #1
You have:
x^4 + 2x^3 - 3x - 6
Take the first two terms and factor out a common x^3:
x^3(x + 2) - 3x - 6
Take the last two terms and factor out a common -3
x^3(x + 2) - 3 (x + 2)
Aha! Now you can factor the common x + 2:
(x^3 - 3)(x + 2)
2007-11-28 17:31:37 · answer #1 · answered by Puzzling 7 · 2⤊ 0⤋
You can't have the -6 inside the parentheses since it's not multiplied by x.
x(x^3 + 2x^2 - 3) - 6.
2007-11-28 17:28:58 · answer #2 · answered by Useless Knowledge Goddess 4 · 0⤊ 0⤋
x^4+2x^3-3x-6
factor x^4+ 2x^3
which is x^3(x+2)
then -3x-6
-3(x+2)
(x^3-3) and (x+2)
2007-11-28 17:33:15 · answer #3 · answered by sizemore609 3 · 0⤊ 0⤋
Sorry but what u did is wrong.
One factor that will go thru is (x + 2)
Then you can get the remaining by using Long division(Division of polynomials)
2007-11-28 17:37:42 · answer #4 · answered by Dike A 1 · 0⤊ 0⤋