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

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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.**

To find the roots of x^3 – 2x^2 – 23x + 60.

Let f(x) = x^3-2x^2-23x+60.

We see that f(-5) = (-5)^3-2(-5)^2-23(-5)+60.

f(-5) = -125-50+115+60

f(-5) = -175+175 = 0.

So f(-5) = 0.

Therefore by remainder theorem, if f(0) = 0, then x-a is a factor of f(x).

So f(x) = x^3 – 2x^2 – 23x + 60 has a factor x-(-5) = x+5

Therefore we can write x^3 – 2x^2 – 23x + 60 = (x+5)(x^2+kx+12) as an identity. Put x = 1 in this equation. 1-2-23+60 = (1+5)(1+k+12).

36 = 6(k+13)

36/6 = (k+13)

6 = k+13

So k = 6-13 = -7

Therefore x^3 – 2x^2 – 23x + 60 = (x+5)(x^2-7x+12)

=> x^3 – 2x^2 – 23x + 60 = (x+5)(x-3)(x-4)

So setting each factor to zero, we get the roots :

x+5 = 0, x-3 = 0, x-4 = 0.

**So x= -5, x= 3 and x= 4 are the roots.**