Let h(x) = 2x^5 - 4x^4 - 6x^3 + 5x^2 + 2x - 3. List the possible zeros according to the Rational Zeros Theorem.

Question

Please answer my question and show me how you got it as soon as you can.

Answer 1

possible rational zeros are integer factors of the constant term over factors of the coefficient of the leading x^5 term

3/2, -3/2, 1/2, -1/2, 1/1=1, -1/1=-1, 3/1=3, or -3/1=-3 are the possible rational roots

integer factors of the constant term are - 1, +1, -3 and +3

integer factors of the coefficient of x^5 are 1, -1, 2 and -2

there may be irrational or imaginary roots too.

3 possible positive roots, 2 possible negative roots. 5 roots. 0, 2 or 4 are imaginary

Decartes' rule of signs, 3 changes in signs of coefficients. substitute -x for x and 2 changes in signs

imaginary roots come in conjugate pairs, always an even number 5 total roots or zeros.

x h(x)

0 -3

1 -4

2 -27

3 48 one zero is between 2 and 3 as h(x) changes sign 2<x<3, closer to 2, but it can't be rational.

-1 -1

-2 -195

-3 even more negative

3 and -3 are not zeros, 1 and -1 are not, although -1 was closest

that leaves 1/2, -1/2, 3/2 and -3/2

none seem to work. It's a little tedious checking them, plugging them into h(x) to see if it =0