How do you find all the zeros of f(x)=x^3 -x^2 -4x -6f(x)=x3−x2−4x−6?

1 Answer
George C.
Apr 1, 2016

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

x=3x=3

x=-1+-ix=−1±i

Explanation:

By the rational root theorem, any rational zeros of this f(x)f(x) must be expressible in the form p/qpq with integers p, qp,q with pp a divisor of the constant term -6−6 and qq a divisor of the coefficient 11 of the leading term.

So the only possible rational roots are:

+-1±1, +-2±2, +-3±3, +-6±6

Trying each in turn, we find:

f(3) = 27-9-12-6 = 0f(3)=27−9−12−6=0

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

x^3-x^2-4x-6 = (x-3)(x^2+2x+2)x3−x2−4x−6=(x−3)(x2+2x+2)

The remaining quadratic factor is of the form ax^2+bx+cax2+bx+c with a=1a=1, b=2b=2 and c=2c=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

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