Question

how to find the zeros of \[x ^{3}\] + x + 10

Answer 1

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

Answer 2

but how to find the first step?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

\[x=\frac{p}{q}=\frac{-2}{1}=-2\]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.

Answer 8

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

Answer 9

long division? hm never heard of that

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

http://www.khanacademy.org/video/polynomial-division?playlist=ck12.org%20Algebra%201%20Examples

Answer 15

ah, okay :)

Answer 16

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

Answer 17

kk thx lucas :)

Answer 18

You're welcome, Markus :)

Answer 19

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).

Answer 20

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

Answer 21

np ;)

Answer 22

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