# Thread: How do I do this?

1. ## How do I do this?

Find all real and imaginary zeros of f(x) = x^3 + x^2 + 13x - 15, given that 1 is a zero of the function.

Would I use synthetic division? If so, how?

Any help is appreciated.

2. Originally Posted by iluvmathbutitshard
Find all real and imaginary zeros of f(x) = x^3 + x^2 + 13x - 15, given that 1 is a zero of the function.

Would I use synthetic division? If so, how?

Any help is appreciated.
you have $f(1)=0$ which means $(x-1)$ is a factor.

3. ok, so I would use that for synthetic division?

4. Originally Posted by iluvmathbutitshard
ok, so I would use that for synthetic division?

i believe yes.
divide $x^3+x^2+13x-15$ by $x-1$.

5. Ok, so when I divided I got x^2 + 2x + 15?
How are these zeros?

6. Originally Posted by iluvmathbutitshard
Ok, so when I divided I got x^2 + 2x + 15?
How are these zeros?
use the ABC formula to find the roots

7. Originally Posted by iluvmathbutitshard
Ok, so when I divided I got x^2 + 2x + 15?
How are these zeros?
"They" aren't zeros. The zeros of the original polynomial are 1 and the zeros of this quadratic polynomial. Solve x^2+ 2x+ 15= 0 by completing the square or the quadratic equation.

8. $x^3 + x^2 + 13x - 15=(x-1)(x^2 + 2x + 15)=(x-1)((x+1)^2+14).$

9. The imaginary zeros are obtained from x^2+2x+15,
since this factor of the cubic does not itself have real factors,
it has complex factors since b^2<4ac.
x^2+2x+15 is zero for x = {-2+sqrt(4-60)}/2
and for x = {-2-sqrt(4-60)}/2,
which are x = -1+i(sqrt[14]) and x = -1-i(sqrt[14]).