# What's a factor of x^3 - x - 6 =0

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We will substitute with x= 2.

==> x^3 - x - 6 = 0

==> 8 - 2 -6 = 0

Then x= 2 is a root for the equation.

==> (x-2) is a factor.

Now we will divide by (x-2) to find other factors.

==> (x^3 - x - 6) = (x-2) (x^2 + 2x+ 3)

Now sine delta = (4-4*3) = 4-12 = -8 < 0

Then there are no Real roots foe the quadratic equation.

**Then the factors of the equation are : (x-2) and (x^2 + 2x+3)**