Question

How many real roots would f(x)=
x^3+2x^2+2x-5 have

Answer 1

One real root

Explanation:

By the fundamental theorem of algebra, an nth degree polynomial has n-possible real roots.

If there are complex roots, then the the complex roots come in pairs.

Therefore the number of possible real roots of
`f(x)=x^3+2x^2+2x-5` are 3 real roots with no complex pairs or 1 real root with a complex pair.

By Descartes rule of signs, there is only one change of sign in the polynomial (+ to -).

Hence there is only one positive real root.

`f(-x)=-x^3+2x^2-2x-5`

There is change is sign two times but we can not have even number of real roots for this polynomial

Therefore there is only real root.