Question

list the possible rational zeros for f(x) = x^4 + 3x^3 - 15x^2 - 19x + 30

Answer 1

That would be the factors of 30 divided by the coefficient of the x^4 term which is 1.

Answer 2

To solve a quartic equation it is often easiest to test to see if 1 or -1 is a root, which is relatively easily done, and then factor to get a cubic. In this case, you can see that 1 is a solution to the equation (because 1 + 3 - 15 - 19 + 30 = 0). That means that (x-1) is a factor, so you can factor out the (x-1) to get \((x-1)(x^3+4x^2-11x-30)\). Then, you can factor the cubic to get the remaining roots, 3, -2, and -5.

Answer 3

so the possible rational zeros is 1?

Answer 4

\[\pm1, pmw2, \pm3, \pm5, \pm6, \pm 10, \pm 15\pm\]

Answer 5

@Mertsj's way sounds easier, I had not known you could do that.