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

1 Answer
Jan 22, 2016

Answer:

#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]}