but i was only able to find two x intercepts 2 and 6/5 wen i know that there is suppose to be a third one, because the function hits the x axis 3 times. Can someone varify my answer and tell me wut's wrong.
2007-09-14 13:37:01 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Product of Roots = -6
Sum of roots = 2
The group that multiplies to
1, 3, -2
So the x-values are
1, 3, -2
_________________________________
I didn't try factoring it because it would be too hard for me.
What I did was guess numbers, but I knew some of the requirements so it was easy for me.
2007-09-14 13:46:37 · answer #1 · answered by UnknownD 6 · 0⤊ 0⤋
x^3 -2x^2 - 5x + 6 = 0
substituting x = -2 the expression is zero
so x+ 2 is a factor
divide the expression by ( x + 2)
x^3 -2x^2 - 5x + 6/(x + 2) = x^2 - 4x + 3
factorise (x ^2 - 4x + 3)
x^2 - 3x - x + 3 = 0
x((x - 3) - 1( x - 3) =0
(x - 3)( x - 1) =0
so x = 3 and x = 1
so x intercepts are 1, - 2 and 3
2007-09-14 13:52:35 · answer #2 · answered by mohanrao d 7 · 0⤊ 0⤋
1 works 1-2-5+6=0
2 doesn't work 8-8-10+6=4
2007-09-14 13:45:08 · answer #3 · answered by batman is a terrible superhero 1 · 0⤊ 0⤋
y=x^3-2x^2-5x+6
By inspection x = 1 is a root so (x-1) is a factor.
Use long division to find (x-1)(x^2-x-6) are factors of y.
So y = (x-1)(x-3)(x+2)
So roots are 1 3, and -2
2007-09-14 13:50:33 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋