anonymous
  • anonymous
Prove that (x-a) is a factor of x^3 - (a + b + c)x^2 + (ab +bc + ca)x-abc. How do I prove this???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
show that if you replace \(x\) by \(a\) you get 0
anonymous
  • anonymous
then you know if \(a\) is a root of a polynomial \(p(x)\) then you can factor as \(p(x)=(x-a)(q(x))\)
anonymous
  • anonymous
yeah i tried that, but i couldnt end up with a zero..

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anonymous
  • anonymous
if you do not get 0, then you cannot factor
anonymous
  • anonymous
i probably made a mistake.. because it has to equal to zero
anonymous
  • anonymous
\[a^3 - (a + b + c)a^2 + (ab +bc + ca)a-abc\] is a startr
anonymous
  • anonymous
yeah it works, try again
anonymous
  • anonymous
a^3 - (a + b +c)a^2 + (ab+ bc +ca)a -abc = a^3 - a^3 + a^3b + a^3 c + a^2b + bc+ca^ - abc
anonymous
  • anonymous
ohhh i see it, i messed up
anonymous
  • anonymous
hmmm forgot the distributive property for the second term
anonymous
  • anonymous
and some other mistakes too, but you can clean it up i am sure
anonymous
  • anonymous
okay i tried and i ended up like this: 2a^2 + 2a^2c
anonymous
  • anonymous
lets go slow
anonymous
  • anonymous
ohh its negative.. i got it
anonymous
  • anonymous
first step \[a^3 - (a + b + c)a^2 + (ab +bc + ca)a-abc\] second step ok
anonymous
  • anonymous
yeah i didnt see the negative sign infront of the first bracket, thank you so much!!
anonymous
  • anonymous
yw (you did all the work)

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