How do you find all the zeros of #f(x)=x^3 -x^2 -4x -6#?

1 Answer
Apr 1, 2016

Answer:

Use the rational root theorem and completing the square to find:

#x=3#

#x=-1+-i#

Explanation:

By the rational root theorem, any rational zeros of this #f(x)# must be expressible in the form #p/q# with integers #p, q# with #p# a divisor of the constant term #-6# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational roots are:

#+-1#, #+-2#, #+-3#, #+-6#

Trying each in turn, we find:

#f(3) = 27-9-12-6 = 0#

So #x = 3# is a zero and #(x-3)# a factor:

#x^3-x^2-4x-6 = (x-3)(x^2+2x+2)#

The remaining quadratic factor is of the form #ax^2+bx+c# with #a=1#, #b=2# and #c=2#. This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 2^2-(4*1*2) = 4-8 = -4#

Since this is negative, the quadratic has no Real zeros and no linear factors with Real coefficients.

We can factor it with Complex coefficients by completing the square:

#x^2+2x+2#

#= x^2+2x+1+1#

#= (x+1)^2-i^2#

#= ((x+1)-i)((x+1)+i)#

#= (x+1-i)(x+1+i)#

Putting it all together:

#x^3-x^2-4x-6#

#= (x-3)(x^2+2x+2)#

#= (x-3)(x+1-i)(x+1+i)#

Hence zeros:

#x=3#

#x=-1+-i#