How do you find all the zeros of f(x)=x3−4x2+14x−20?

1 Answer
George C.
Jun 5, 2016

x=2 or x=1±3i

Explanation:

f(x)=x3−4x2+14x−20

By the rational root theorem, any rational zeros of f(x) must be expressible in the form pq for integers p,q with p a divisor of the constant term −20 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are:

±1, ±2, ±4, ±5, ±10, ±20

In addition note that if you invert the signs of the coefficients of the terms of odd degree, then the pattern of signs you get is:

−−−−

which has no changes. So there are no negative zeros.

So the only possible rational zeros are:

1,2,4,5,10,20

We find:

f(2)=8−16+28−20=0

So x=2 is a zero and (x−2) a factor:

x3−4x2+14x−20=(x−2)(x2−2x+10)

The remaining quadratic factor has negative discriminant, so no Real zeros, but we can still factor it by completing the square and using the difference of squares identity:

a2−b2=(a−b)(a+b)

with a=(x−1) and b=3i as follows:

x2−2x+10

=x2−2x+1+32

=(x−1)2−(3i)2

=((x−1)−3i)((x−1)+3i)

=(x−1−3i)(x−1+3i)

Hence Complex zeros:

x=1±3i

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