how to solve P(x)=x^3-5x^2+2x+8

Question

find all rational roots for P(x)=0.

Answer 1

If a, b, c are the roots then (x-a)(x-b)(x-c)=x³-5x²-2x+8. The constant term is -abc=8. 

The unsigned rational roots are x=1, 2, 4, because 1×2×4=8, the constant term. We can have:

-abc=(-1)(2)(4)=(1)(-2)(4)=(1)(2)(-4)=8. In other words, one root must be negative and the other two positive.

In actual fact, -1 is a true root, because -1-5-2+8=0. Therefore, the other rational roots can only be 2 and 4. Knowing x=-1 is a root we can use synthetic division to reduce the cubic to a quadratic:

-1 | 1 -5 2    8

      1 -1 6 | -8

      1 -6 8 |  0 =x²-6x+8=(x-4)(x-2), so the other true roots are 2 and 4, which are also the rational roots.