anonymous
  • anonymous
how to find the zeros of \[x ^{3}\] + x + 10
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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nikvist
  • nikvist
\[x^3+x+10=0\] \[(x+2)\cdot(x^2-2x+5)=0\] \[(x+2)\cdot((x-1)^2+4)=0\] \[x_1=-2\quad,\quad x_{2,3}=1\pm 2i\]
anonymous
  • anonymous
but how to find the first step?
anonymous
  • anonymous
well i see - something you have to get used to ^^ thanks a lot

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anonymous
  • anonymous
Neasd, you can use the 'rational root theorem' to test for rational roots.
anonymous
  • anonymous
The roots will be of the form\[\frac{p}{q}=\frac{\pm \left\{ factors.of.constant.term \right\}}{\left\{ factors.of.coefficient .of.highest.power \right\}}\]
anonymous
  • anonymous
So here,\[\frac{p}{q}=\frac{\pm \left\{ 1,2,5,10 \right\}}{\left\{ 1 \right\}}\]
anonymous
  • 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.
anonymous
  • anonymous
Do you know how to use long division on polynomials? That is the next step.
anonymous
  • anonymous
long division? hm never heard of that
anonymous
  • anonymous
The aim is to find Q(x) such that\[x^3+x+10=(x+2)Q(x)\]
anonymous
  • anonymous
It's a little difficult to explain on this site. I'll see if I can find a clip that will show you.
anonymous
  • anonymous
a i see polynomdivision ;D my first language is german ^^
anonymous
  • anonymous
yeah i know to do that :D thank you!!
anonymous
  • anonymous
http://www.khanacademy.org/video/polynomial-division?playlist=ck12.org%20Algebra%201%20Examples
anonymous
  • anonymous
ah, okay :)
anonymous
  • anonymous
The division will find the Q(x), which is the quadratic Nikvist found.
anonymous
  • anonymous
kk thx lucas :)
anonymous
  • anonymous
You're welcome, Markus :)
anonymous
  • 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).
anonymous
  • anonymous
ok now i'm prepared :D thx again ;)
anonymous
  • anonymous
np ;)
anonymous
  • anonymous
may i ask you, what you are doing atm? student? postdoc etc? :D

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