Question

Given -2/3 is a zero of f(x)=3x^3-4x^2+8x+8, find all remaining zeros.

Answer 1

\[f(x)=3x^3-4x^2+8x+8,\]

the possible zeros are\[±1,~~±2,~~±4,~~±8.\]

Answer 2

@m.Rebecca That first response is wrong. I'll work to figure out the right answer.

Answer 3

hmm...can I ask why?

Answer 4

It's a cubic function, so the most zeros it can have is 3.

Answer 5

I am not saying all of them work, but these are the +- factors of the constant (or 8)

Answer 6

You should plug them in and see which of them work.

Answer 7

If you are accounting for nonreal solutions, then \[1\pm i \sqrt{3}\] are 2 other solutions.

Answer 8

You are supposed to explain your work @alex10101.

Answer 9

okay I get the answer, but I don't understand how

Answer 10

divide f(x) by x-(-2/3)
then you have quadratic

Answer 11

The clue here is that we know (x+2/3) must be a factor of that function because -2/3 is a zero. If we eliminate this factor from the original polynomial, we'll be left with a quadratic:
\[\frac{ 3x ^{3}-4x ^{2}+8x+8 }{ x+2/3 }=3x ^{2}-6x+12\]
From here we can use the quadratic equation to find the 2 imaginary solutions:
\[x=\frac{ -(-6)\pm \sqrt{(-6)^{2}-4(3)(12)} }{ 2(3) }=\frac{ 6\pm \sqrt{-108} }{ 6 }=\frac{ 6\pm 6\sqrt{-3} }{ 6 }=1\pm i \sqrt{3}\]
Hopefully this helped clear up some confusion.

Answer 12

@amoodarya's recommendation result.
3 x^2 - 6 x + 12

Answer 13

@robtobey what?

Answer 14

(3 x^3 - 4 x^2 + 8 x + 8)/(x+(2/3)) = 3 x^2 - 6 x + 12
Solve 3 x^2 - 6 x + 12 = 0 for x.

Answer 15

but how do I get the 3x^2-6x+12?
like I can set it up but I don't know how to solve it

Answer 16

Honestly that part I just used my calculator. It's not easy to recognize any factors of a cubic, but it is possible to use long-division of polynomials to find how much is left when you divide (3x^3-4x^2+8x+8) by (x+2/3).

Answer 17

Here's a website that explains polynomial long-division:
http://www.purplemath.com/modules/polydiv2.htm

Answer 18

Thank You!

Answer 19

No probs