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...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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.**