# Thread: Use the given zero to find the remaining zeros of each polynomial function

1. ## Use the given zero to find the remaining zeros of each polynomial function

Finding the zero's of P(x) 3x^3-29x^+92x+34. The given zeros are 5+3i

2. Originally Posted by renebran
Finding the zero's of P(x) 3x^3-29x^+92x+34. The given zeros are 5+3i
If $\displaystyle 5 + 3i$ is a zero, then so is $\displaystyle 5 - 3i$.

So $\displaystyle x - (5 + 3i)$ and $\displaystyle x - (5 - 3i)$ are factors.

Therefore so is $\displaystyle [x - (5 + 3i)][x - (5 - 3i)]$

$\displaystyle = x^2 - x(5 - 3i) - x(5 + 3i) + (5 + 3i)(5 - 3i)$

$\displaystyle = x^2 - 5x + 3ix - 5x - 3ix + 25 + 9$

$\displaystyle = x^2 - 10x + 34$.

So divide your polynomial by $\displaystyle x^2 - 10x + 34$ in order to find the third factor (and third zero).

3. Originally Posted by Prove It
If $\displaystyle 5 + 3i$ is a zero, then so is $\displaystyle 5 - 3i$.

So $\displaystyle x - (5 + 3i)$ and $\displaystyle x - (5 - 3i)$ are factors.

Therefore so is $\displaystyle [x - (5 + 3i)][x - (5 - 3i)]$

$\displaystyle = x^2 - x(5 - 3i) - x(5 + 3i) + (5 + 3i)(5 - 3i)$

$\displaystyle = x^2 - 5x + 3ix - 5x - 3ix + 25 + 9$

$\displaystyle = x^2 - 10x + 34$.

So divide your polynomial by $\displaystyle x^2 - 10x + 34$ in order to find the third factor (and third zero).
This is where I'm having the problem. I'm not sure how to divide the polynomial .

4. Originally Posted by renebran
This is where I'm having the problem. I'm not sure how to divide the polynomial .
Use long division.