^{2}-1. If you perform the division, you will get:

^{3}-8x

^{2}-x+2=0

^{3}-8x

^{2}-x+2 by (2x-1), you will get 2x

^{2}-3x-2=0 equation. This can be solved using quadratic formula. It yields:

_{1}=[3+√25]/4=2;

_{2}=[3-√25]/4=-1/2;

i dont how to get the Q from P/Q to find the possible zero

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I would try to see which interval/intervals the roots may lie into. First, I would see what I get if I plug in x=0, I will get -2; If I plug in -1, I will get zero, which means x=-1 is the root (lucky me!). It turns out that x=1 is also the root. Thus you can now divide your polynomial by (x-1)*(x+1)=x^{2}-1. If you perform the division, you will get:

4x^{3}-8x^{2}-x+2=0

This equation at x=0 yields 2. At x=1 it gives -3; at x=-1 it gives -9. Thus, there are two more roots, one in the interval [-1;0], another in the interval [0;1]. Since the product of all roots must be -2/4=-½ and we already have two roots, 1 and (-1), it is reasonable to try 1/2. Indeed, this is the root. Let us divide 4x^{3}-8x^{2}-x+2 by (2x-1), you will get 2x^{2}-3x-2=0 equation. This can be solved using quadratic formula. It yields:

x_{1}=[3+√25]/4=2;

x_{2}=[3-√25]/4=-1/2;

Thus you have the following roots:

-1; -1/2; 1/2; 1; 2

factors of -2 are -1, 2, 1, -2

factors of 4 are 1, 4, 2, -1, -4, -2

possible zeros are -1, 1, 1/2, -1/2

try 1

divide the polynomial by (x - 1) and see if the remainder is zero.

Or simply plug in the value 1.

If it is, then 1 is a zero or root of f(x) = 0

If remainder is not zero, try another possibility from the choices

It turns out 1 is a zero.

When you do divide and get the quotient which is a 4th degree polynomial, you can continue the same process to derive additional zeros.

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