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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
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show that if you replace \(x\) by \(a\) you get 0
then you know if \(a\) is a root of a polynomial \(p(x)\) then you can factor as \(p(x)=(x-a)(q(x))\)
yeah i tried that, but i couldnt end up with a zero..

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Other answers:

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

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