im confused im supposed to factor the polynomial and find the zeros...yet i dont know why im wrong...help someone please!
f(x)=(3x^3)-(x^2)-12x+4
this is what i have :(x+2)(x-2)(3x+1)(3x-1)
and the zeros are: -2,2,-1/3,1/3
2007-09-30 07:41:42 · 6 answers · asked by ...i <3 Xavi... 2 in Science & Mathematics ➔ Mathematics
Your only problem is that you have an extra factor thrown in there. It really factors out into : f(x)=(3x-1)(x+2)(x-2)
So the zeros end up being: 1/3, -2, & 2
2007-09-30 07:44:12 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Factor by grouping...
3x^3 - x^2 - 12x + 4
x^2 (3x - 1) - 4 (3x - 1)
Factor out the common factor of (3x - 1)...
(3x - 1)(x^2 - 4)
Factor the difference of squares...
(3x - 1)(x + 2)(x - 2)
Set it equal to zero...
3x - 1 = 0 ..or.. x + 2 = 0 ..or.. x - 2 = 0
x = 1/3 ..or.. x = -2 ..or.. x = 2
Your error was in the first step. When you factored -4 from the second pair, you mistakenly got (3x + 1), which did not provide a common factor for the second step. Remember, when the highest power is x^3, there are only 3 answers.
2007-09-30 15:11:19 · answer #2 · answered by JM 4 · 0⤊ 0⤋
First of all, you can only have as many factors as the highest degree of X is. You should have 3 factors.
Synthetic division brings you to f(x)=(3x^2+5x-2) (x-2)
Quadratic formula makes it (3x-1)(x+2) (x-2)
2007-09-30 14:57:03 · answer #3 · answered by Sith Lord 13 2 · 0⤊ 0⤋
A third degree equation can´t have four roots
f(x) = x-2)( 3x^2-5x-2)
and 3x^2-5x-2=0
x= ((5+-sqrt(49)/6 = 2 and -1/3
sof(x)=3(x-2)^2*(x+1/3)
2007-09-30 14:50:26 · answer #4 · answered by santmann2002 7 · 0⤊ 1⤋
x²(3x-1)-4(3x-1)
(x²-4)(3x-1)
x=2,-2,1/3
Third degree means 3 roots most
You didn't factor right.
2007-09-30 14:46:57 · answer #5 · answered by chasrmck 6 · 0⤊ 1⤋
A cubic equation can not have four roots
2007-09-30 14:47:19 · answer #6 · answered by fcas80 7 · 0⤊ 1⤋