anonymous
  • anonymous
What is the factor of x^3-4x^2+x+6?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
here is a good gimmick to know: it is going to be easy to factor if you can find the zeros math teachers usually make a zero of 1 or -1 because those are the easiest, so check those first
anonymous
  • anonymous
1 does not work, because if you plug it in to \(x^3-4x^2+x+6\) you get \[1-4+1+6\] which is not zero
anonymous
  • anonymous
but \(-1\) does work, you get \[-1-4-1+6=0\]

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anonymous
  • anonymous
therefore you know it factors \[x^3-4x^2+x+6=(x+1)(something)\] and you can find the "something" either by division (synthetic division is easiest) or by thinking
anonymous
  • anonymous
your choice
anonymous
  • anonymous
What divisor can i use in synthetic division?
anonymous
  • anonymous
you want to factor \(x^3-4x^2+x+6\) as \((x+1)(whatever)\) this makes \[whatever=(x^3-4x^2+x+6)\div (x+1)\]
anonymous
  • anonymous
you can also just figure out what it has to be in your head more or less
anonymous
  • anonymous
i got 3x
anonymous
  • anonymous
how? \[x^3-4x^2+x+6=3x(x-1)?\] that can't be correct, can it?
anonymous
  • anonymous
No, it's wrong. :3
anonymous
  • anonymous
actually i meant to write \[x^3-4x^2+x+6=3x(x+1)?\] but that is not correct either your "whatever" needs to be a polynomial of degree 2
anonymous
  • anonymous
it is clear what you have to do? you have to write \[x^3-4x^2+x+6=(x+1)(ax^2+bx+c)\]

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