Possible rational zeroes for polynomial

F(x)=3x^3+3x^2+2x-6

I didn't find any rational zeros. I know it crossed the y-axis somewhere, but can't factor out the zeros.

woops crossed X-axis somwehere.

RATIONAL ROOTS TEST

Just because it crosses the X-axis does not mean this has rational roots.

Anyway, maple solve this and so the only real root is : 1/3 (29+sqrt(842))^(1/3)-(1)/(3 (29+sqrt(842))^(1/3))-1/

Sheeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeiit.

lolz... POSSIBLE.

read the question again OP.

>>5
haha, good eye

RATIONAL ROOTS TEST GOD DAMN IT

>>5
>>6
wat

>>7
Rational roots test is useless.

(x-1/2)(x-2) = x^2 - 5/2x + 1

But rational roots test gives the only possible rational solution as x = 1, so obviously doesn't give all of them.

>>9
That's because the rational root test only applies to polynomials with integer co-efficients, idiot.

If you understood why it worked (Which is fucking obvious, by the way), instead of just blindly applying it you'd be in a lot better stead.

The polynomial you want is (2x-1)(x-2)= 2x^2 - 5x + 2


Want a proof of the rational root test?

Say f(p/q)=0 where f(x)=sum a_i.x^i, f has degree n.

then multiply through by q^n.

q^n.f(p/q) = sum a_i.p^i.q^n-i = 0

But q divides a_i.p^i.q^n-i for all i != n, so this implies q divides a_n.p^n, but since p/q is reduced q divides a_n.


Similarly p divides a_i.p^i.q^n-i for all i != 0, so this implies p divides a_0.q^n, so p divides a_0.