anonymous
  • anonymous
How to factor: x^5-6x^4+40x^2-13x-70 Please show me the steps!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
There is a theorem thing that states that plus or minus the factors of the constant(70 in this case) divided by plus or minus the constant in front of the term with the largest power are all the possibilities of the factors of the polynomial. So just start crunching the numbers.
anonymous
  • anonymous
try \[\pm1,2,5,7,10,35,70\]then use synthetic division or polynomial long division to divide out the factors
anonymous
  • anonymous
can you explain synthetic divsion please?

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anonymous
  • anonymous
let finish solving the problem and then I will explain
anonymous
  • anonymous
Then, can you please show the steps for doing the polynomial long division to divide out the factors
anonymous
  • anonymous
there is only one integer factor: (x-2). |dw:1357004129003:dw|
anonymous
  • anonymous
the other factor is (x-2)
anonymous
  • anonymous
What are the numbers of the thrid and forth row?
anonymous
  • anonymous
oh, I am beginning to see how synethetic divsion works
anonymous
  • anonymous
Please correct me if I am going to wrong way: x^5-6x^4+40x^2-13x-70 =(x-2)(x^4-4x^3-8x^2+24x+35) The next step would be to divide (x^4-4x^3-8x^2+24x+35) by 5 using synethetic divsion?
anonymous
  • anonymous
The remaining factors are not rational, so the rational roots theorem would not assist us any further.
anonymous
  • anonymous
oh, no wonder it didn't work anymore
anonymous
  • anonymous
so that's how you factor it
anonymous
  • anonymous
(x-2)(x^4-4x^3-8x^2+24x+35) This is the simplest form? Can't (x^4-4x^3-8x^2+24x+35) be factor too?
anonymous
  • anonymous
Since the remaining roots are complex, you couldn't factor out a single term, you would have to somehow factor out a quadratic. I do not know any techniques that will do this.
anonymous
  • anonymous
Not complex, irrational. But they exist as conjugates so you would have to factor out a quadratic instead of a single term.
anonymous
  • anonymous
ok, i will continue to try
anonymous
  • anonymous
I got it (x^4-4x^3-8x^2+24x+35)=(x^2-2x-7)(x^2-2x-5)
anonymous
  • anonymous
so when (x^5-6x^4+40x^2-13x-70) is factored, it will be (x-2)(x^2-2x-7)(x^2-2x-5) Is it correct?
anonymous
  • anonymous
Yes. That's it !
mathmate
  • mathmate
@yociyoci just curious, how did you factorize it?
anonymous
  • anonymous
Thank you so much everyone for helping me with factoring! I learnt so much from you guys!! Thanks!!
anonymous
  • anonymous
I plug in variables and did some algebra and trial and error
anonymous
  • anonymous
o_o
anonymous
  • anonymous
ya, it was long...
mathmate
  • mathmate
You can also use some facts as we usually do in quadratics. Assume (x^2+ax+b)(x^2+cx+d) then (a+c)=-4 ac+b+d=-8 bc+ad=24 bd=35 Hopefully this will help you eliminate cases quickly without having to multiply out the whole expression.
anonymous
  • anonymous
ya, I did something similar
anonymous
  • anonymous
That's a really good technique.
mathmate
  • mathmate
Great! Thank you for the enjoyable session!
anonymous
  • anonymous
Thank you very much, guys!! :)

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