Thread: Determining all the zeros of a function

1. Determining all the zeros of a function

Determine the zero's

f(x) = kx^3 - 8x -x +3k + 1 k=3 and has a zero when x = 2

3x^3 - 8x^2 - x + 3(3) + 1

= 3x^3 - 8x^2 - x +10 <<<<< don't know what to do after this step

2. Originally Posted by Devi09
...
It's hard to follow what you did: what happened to k?

3. Well you had to first find the value of k which is 3 after you solve for it given one of the zeros which was 2

4. You probably have to factor 3x^3 - 8x^2 - x +10 to get the zeros. But Im stuck at that part

5. Originally Posted by Devi09
You probably have to factor 3x^3 - 8x^2 - x +10 to get the zeros. But Im stuck at that part
You know that x=2 is a root, so (x-2) is a factor. So divide 3x^3 - 8x^2 - x +10 by (x-2) to get the quadratic which will give you the two remaining roots.

CB

6. Hello, Devi09!

$\text{Determine the zeros:}$

. . $
f(x) \:=\: kx^3 - 8x^2 -x +3k + 1,\;k=3$

Why do they do that? . . . Can't they plug in the 3?

. . $\text{and has a zero when }x = 2.$

So we have: . $f(x) \:=\:3x^3 - 8x^2 - x + 10$

Since $f(2) = 0$, then $(x\!-\!2)$ is a factor of $\,f(x).$

We find that: . $f(x) \;=\;(x-2)(3x^2-2x-5) \;=\;(x-2)(x+1)(3x-5)$

Therefore, the zeros are: . $x\:=\:2,\,\text{-}1,\,\frac{5}{3}$

Edit: Too slow . . . again.