this is a mega problem
2007-01-17 15:57:11 · 8 answers · asked by ? 1 in Science & Mathematics ➔ Mathematics
it is a polynomial function
2007-01-17 15:59:23 · update #1
Notice that the coefficients are in the same ratio. This means you can factor by grouping:
x³(2x - 1) + (2x - 1)
(x³ + 1)(2x - 1)
And then you can factor the sum of cubes:
(x + 1)(x² - x + 1)(2x - 1)
And that's as far as you can take it.
2007-01-17 16:01:12 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
put x=1 in the above equation that is 2(1)^4 -(1)^3 +2(1)-1=2-3+2-1=0 so we can takex-1 is a factor of above proble.
x-1)2x^4-x^3+2x-1(2x^3+x^2+x+1
2x^4-2x^3
- +
------------------------
0+x^3+2x-1
x^3-x^2
- +
-------------------------
0+x^2+2x-1
x^2+x
- -
---------------------------
0+x-1
x-1
- +
---------------------
0
so factors are(x-1)(2x^3+x^2+x+1)
2007-01-17 20:18:02 · answer #2 · answered by surekha a 1 · 0⤊ 0⤋
2x^4 - x^3 + 0x^2 + 2x - 1
(2x^4 + 2x^3) - (3x^3 - 3x^2) + (3x^2 + 3x) - (x - 1)
2x^3(x+1) - 3x^2(x+1) + 3x(x+1) - 1(x+1)
(x+1)(2x^3 - 3x^2 + 3x - 1)
2007-01-17 17:55:31 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2x^4 - x^3 + 0x^2 + 2x - 1
(2x^4 + 2x^3) - (3x^3 - 3x^2) + (3x^2 + 3x) - (x - 1)
2x^3(x+1) - 3x^2(x+1) + 3x(x+1) - 1(x+1)
(x+1)(2x^3 - 3x^2 + 3x - 1)
2007-01-17 16:19:03 · answer #4 · answered by ? 1 · 0⤊ 0⤋
x^3 *2x -x^3 + 2x-1
x^3 (2x -1) + (2x-1)
(x^3 + 1) * (2x - 1)
(x + 1) * (x^2 - x + 1) * (2x - 1)
2007-01-17 18:05:07 · answer #5 · answered by coollovablechap 1 · 0⤊ 0⤋
it is not a problem. you are missing something
if f(x)=0, then x=1 and x=-1
....
2007-01-17 16:14:03 · answer #6 · answered by mankind 3 · 0⤊ 0⤋
the answer is 0
2007-01-17 16:05:27 · answer #7 · answered by Anonymous · 0⤊ 0⤋
x=-1 or .5, TI-89's kick ***.
2007-01-17 16:02:22 · answer #8 · answered by RH (a.k.a. God) 3 · 0⤊ 0⤋