How do you find all the zeros of #f(x) = x^3 - 4x^(2) + 14x - 20#?

1 Answer
George C.
Jun 5, 2016

#x = 2# or #x = 1+-3i#

Explanation:

#f(x) = x^3-4x^2+14x-20#

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# 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:

#x^3-4x^2+14x-20 = (x-2)(x^2-2x+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:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-1)# and #b=3i# as follows:

#x^2-2x+10#

#=x^2-2x+1+3^2#

#=(x-1)^2-(3i)^2#

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

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

Hence Complex zeros:

#x = 1+-3i#

Related questions
See all questions in Zeros
Impact of this question
5741 views around the world
You can reuse this answer
Creative Commons License