## anonymous 5 years ago any help would be appreciated.. I have a test, and I am unsure of my probs below... ty.

1. anonymous

where are the problems?

2. anonymous

I have one... Given that the polynomial f(x) has a zero at x=2 find the remaining zeros.. f(x)=x^3-10x^2+41x-50

3. nowhereman

You can factor (x-2) from f(x) if it has a zero at x=2.

4. anonymous

What does that mean? It says my answer should be in a+bi form?

5. anonymous

having a zero at x=2 means that (x-2) is one factor of f(x), you have to find the other factors.. using long division gives: $f(x)=(x-2)(x^2-8x+25)=0$ you already have the solution x=2 to find the other solutions solve the quadratic equation: $x^2-8x+25=0$

6. anonymous

using the quadratic formula gives us the two remaining solutions: $x=4-3i, x=4+3i$

7. anonymous

do you know how to use the quadratic formula?

8. anonymous

Long division.. we are being taught syntheic.. is that ok?

9. nowhereman

there is only one solution to that division, and that step is really trivial it does not matter how you solve it.

10. anonymous

Can you show me the way the quadratic formula is worked out on here??

11. nowhereman

The solution formula for $x^2 + px + q = 0$ is $x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} - q}$

12. anonymous

13. anonymous

becasue I have never seen that before??

14. nowhereman

Some schools teach the form $ax^2 + bx +c = 0$ where you would of course get a formula which is a bit different. But the above is the exact solution formula for such quadratic problems.

15. anonymous

Thanks