What are the roots of x^3 – 2x^2 – 23x + 60?

We have to find the roots of : x^3 – 2x^2 – 23x +
60

x^3 – 2x^2 – 23x + 60 =
0

As the equation has a highest power of x, we will have 3
roots.

From the numeric term equal to 60 we know that the
product of the roots is 60, substituting values 1,-1, 2, -2 and 3,  we get that 3 is a
root. So now we have to factor out x – 3, the other roots have a product of
20.

x^3 – 2x^2 – 23x + 60 =
0

(x^3 + x^2 – 20x – 3x^2 – 3x + 60) =
0

(x – 3)(x^2 + x – 20) =0

(x
– 3)( x^2 + 5x – 4x – 20) = 0

We can factorize the
quadratic term as 5* -4 = -20 and 5 - 4 = 1

(x – 3)( x(x +
5) – 4(x + 5)) = 0

(x – 3)(x – 4)(x + 5) =
0

This gives the roots of x^3 – 2x^2 – 23x + 60 as x = 3 ,
x = 4 and x = -5.

The required roots are

x = 3 , x = 4 and x =
-5.