anonymous
  • anonymous
Factor: x^4-2x^3+x^2+8x-5 Try it manually...how is it done?
Mathematics
katieb
  • katieb
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ParthKohli
  • ParthKohli
Use the rational roots test?
anonymous
  • anonymous
try it
ParthKohli
  • ParthKohli
Okay. The possible rational roots are: 5 -5 1 -1

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ParthKohli
  • ParthKohli
Checking all of them. Gimme a sec.
hartnn
  • hartnn
forget 1,-1
ParthKohli
  • ParthKohli
Yup, not 1 and -1.
ParthKohli
  • ParthKohli
not 5.
hartnn
  • hartnn
no rational roots
anonymous
  • anonymous
yeah, but it can still be factorised
anonymous
  • anonymous
into two quadratics... what's the point here?
anonymous
  • anonymous
Well...you're factoring it I guess you could get the roots But how do you get to the quadratics?
ParthKohli
  • ParthKohli
Yeah, we can.
hartnn
  • hartnn
i may suggest a way, (x^2+ax+b)(x^2+cx+d) = x^4-2x^3+x^2+8x-5 and compare the co-efficients, 4 unknowns
hartnn
  • hartnn
like bd=-5
ParthKohli
  • ParthKohli
Like for example, we have \(\sqrt2{}\) as a root, then a factor would be \(\rm x^2 - 2\).
anonymous
  • anonymous
Well, @hartnn 's way is very good to get the quadratics

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