Question

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???

Answer 1

show that if you replace \(x\) by \(a\) you get 0

Answer 2

then you know if \(a\) is a root of a polynomial \(p(x)\) then you can factor as \(p(x)=(x-a)(q(x))\)

Answer 3

yeah i tried that, but i couldnt end up with a zero..

Answer 4

if you do not get 0, then you cannot factor

Answer 5

i probably made a mistake.. because it has to equal to zero

Answer 6

\[a^3 - (a + b + c)a^2 + (ab +bc + ca)a-abc\] is a startr

Answer 7

yeah it works, try again

Answer 8

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

Answer 9

ohhh i see it, i messed up

Answer 10

hmmm forgot the distributive property for the second term

Answer 11

and some other mistakes too, but you can clean it up i am sure

Answer 12

okay i tried and i ended up like this: 2a^2 + 2a^2c

Answer 13

lets go slow

Answer 14

ohh its negative.. i got it

Answer 15

first step
\[a^3 - (a + b + c)a^2 + (ab +bc + ca)a-abc\] second step
ok

Answer 16

yeah i didnt see the negative sign infront of the first bracket, thank you so much!!

Answer 17

yw (you did all the work)