Question

How do you find all the real and complex roots of `x^3 - 2x^2 + 10x + 136 =0`?

Answer 1

`x=-4,3+-5i`

Explanation:

We should first try to find a real root of the polynomial. Any real root, since the coefficient of the high highest degree term is `1`, will be a factor of the final term, `136`.

Thus, our potential factors are `+-1,+-2,+-4,+-8,+-17,+-34,+-68,+-136`.

Attempt synthetic division or polynomial long division with the factors until you find one that doesn't give a remainder.

The only factor that will work is `x+4`, which means the polynomial has a root of `color(red)(-4)`.

We can determine the other two roots by dividing the original function and `x+4` to see the remaining quadratic.

`(x^3-2x^2+10x+136)/(x+4)=x^2-6x+34`

To determine the remaining two roots, we can apply the quadratic formula to `x^2-6x+34`.

`x=(-b+-sqrt(b^2-4ac))/(2a)=(6+-sqrt(-100))/2=color(red)(3+-5i`

The graph should have one real root at `-4` and two imaginary roots (it will only cross the `x` axis once):

graph{x^3-2x^2+10x+136 [-15, 15, -2000, 2000]}