Question

Let P(x)=x^3-6x+k. If the sum of the two roots of P(x) is -2, then the third root is...

Answer 1

r1 + r2 = -2, so you would say r2 = -r1 - 2

Then you have
(x-r1)(x+r1+2)(x-r)

Answer 2

where r is 3rd root.

Answer 3

Should I be multiplying this out?

Answer 4

Yes.

Answer 5

no short cuts?

Answer 6

Wolfram Alpha

Answer 7

ye gods. Doing this during a test.

Answer 8

You could so
(x-a)(x-b)(x-c) and then do the b = -a-2 substitution after you are done.

Answer 9

do^

Answer 10

I'm halfway already. Canceling things out.

Answer 11

x^3+(-r1-r+2)x^2+2(-r1+1)x+r1r(2+r1)

Answer 12

You have three equations now.

(-r1-r+2) = 0
2(-r1+1) = -6
r1r(2+r1) = k

Answer 13

Solving this makes me very happy, but are you /absolutely/ sure there isn't a shorter way?

Answer 14

There could be a shorter way, but I don't know it. I like to use the basics to do these things.

Answer 15

Alrighty.

Answer 16

I found the better solution: Vieta's Formula
Let P(x)=x^3-6x+k. If the sum of the two roots of P(x) is -2, then the third root is...
r1r2r3=k
r1r2+r1r3+r2r3=-6
r1+r2+r3=0

r1+r2=-2
-2+r3=0
r3=2