Rational Zeros Math Problem?

Question

h(x) = x^4 + 2x^3 - 23x^2 + 2x - 24

Find ALL RATIONAL ZEROS of the function.

Answer 1

Looks like a job for factoring!

0 = x^4 + 2x^3 - 23x^2 + 2x - 24

Notice the 24 and 23. I figure, let's add 1 to both sides, so we might be able to group:

1 = x^4 + 2x^3 - 23x^2 + 2x - 23

Let's group the ones with -23 coefficient, which leaves x^2+1. Another pair that factors out x^2+1 would be: 2x^3+2x.

x^4+2x(x^2+1)-23(x^2+1) =x^4+(x^2+1)(2x-23)=1

Let's bring the 1 back over and factor some more:

(x^4-1)+(x^2+1)(2x-23) =
(x^2+1)(x^2-1)+(x^2+1)(2x-23) =
(x^2+1)((x^2-1)+(2x-23)) =
(x^2+1)(x^2+2x-24) =
(x^2+1)(x-4)(x+6) = 0

x^2+1 produces no rational zeroes, as the solution to x^2+1=0 is +/- i (x^2+1=0 -> x^2=-1 -> x = +/- i). Thus, we see the other two roots, being 4 and -6.

Answer 2

x=-6
x=4

Answer 3

^^ And they're both double zeros.