## anonymous 5 years ago how to find the zeros of $x ^{3}$ + x + 10

1. nikvist

$x^3+x+10=0$ $(x+2)\cdot(x^2-2x+5)=0$ $(x+2)\cdot((x-1)^2+4)=0$ $x_1=-2\quad,\quad x_{2,3}=1\pm 2i$

2. anonymous

but how to find the first step?

3. anonymous

well i see - something you have to get used to ^^ thanks a lot

4. anonymous

Neasd, you can use the 'rational root theorem' to test for rational roots.

5. anonymous

The roots will be of the form$\frac{p}{q}=\frac{\pm \left\{ factors.of.constant.term \right\}}{\left\{ factors.of.coefficient .of.highest.power \right\}}$

6. anonymous

So here,$\frac{p}{q}=\frac{\pm \left\{ 1,2,5,10 \right\}}{\left\{ 1 \right\}}$

7. anonymous

$x=\frac{p}{q}=\frac{-2}{1}=-2$is one element of the set. When you test this, you find$(-2)^3+(-2)+10=-8-2+10=0$so x=-2 is a root. You can then factor $(x-(-2))=(x+2)$out of your polynomial.

8. anonymous

Do you know how to use long division on polynomials? That is the next step.

9. anonymous

long division? hm never heard of that

10. anonymous

The aim is to find Q(x) such that$x^3+x+10=(x+2)Q(x)$

11. anonymous

It's a little difficult to explain on this site. I'll see if I can find a clip that will show you.

12. anonymous

a i see polynomdivision ;D my first language is german ^^

13. anonymous

yeah i know to do that :D thank you!!

14. anonymous
15. anonymous

ah, okay :)

16. anonymous

The division will find the Q(x), which is the quadratic Nikvist found.

17. anonymous

kk thx lucas :)

18. anonymous

You're welcome, Markus :)

19. anonymous

Just want to add, if you test all the possible rational combinations and none of them satisfy the condition of a root (i.e. your polynomial does not go to zero), the polynomial has *no* rational roots (you're then left with irrational or complex).

20. anonymous

ok now i'm prepared :D thx again ;)

21. anonymous

np ;)

22. anonymous

may i ask you, what you are doing atm? student? postdoc etc? :D