Question

Determine the zeros of f(x) = x^3 – 12x^2 + 28x – 9.

Answer 1

Do we know about the rational zeros theorem at all?

Answer 2

Yes

Answer 3

Okay, cool. So we need to use that to see what our possible zeros could be. If you know what that is then I'll just go through it.

\[possible zeros = \frac{ \pm9, 3, 1 }{ \pm1 }\]. So know we need to choose one of these possible numbers and see if we can then do synthetic division and come up with a remainder of zero. Do we know how to do synthetic division?

Answer 4

yes

Answer 5

Do i test all of the factors now? If so I got, \[\pm1,\pm3,\pm9\]

Answer 6

Correct. We would need to test those out and hope we get a remainder of 0. I would try positive 9 first :P

Answer 7

So my equation now is \[x ^{2}-3x+1\]?

Answer 8

Now i find the discrimant and factor right?

Answer 9

Correct. Since you got a remainder of zero, this means (x-9) is a factor. And yes, you would then need to factor that portion.

Answer 10

so what would my answer be?

Answer 11

Did you try to factor it?

Answer 12

I didn't get real rumbers, what did i do wrong?

Answer 13

Well, I did it using the quadratic formula, so let me work it out for ya that way

Answer 14

\[\frac{ -(-3)\pm \sqrt{(-3)^{2}-4(1)(1)} }{ 2 }\]
\[\frac{ 3\pm \sqrt{5} }{ 2 }\]
So no, you should get real numbers.

Answer 15

so what would be my solution?

Answer 16

\[(x-9) (x -\frac{ 3+\sqrt{5} }{ 2 }) (x-\frac{ 3 - \sqrt{5} }{ 2 })\] That would be how I'd personally put it. Unless you need to actually solve for x of course.

Answer 17

So my zeroes are those answers?

Answer 18

Yes, those would be your zeros.

Answer 19

thanks