How do you find the zeros of #f(x) = x^3 − 12x^2 + 47x − 60#?

1 Answer
Jun 15, 2016

Answer:

#x = 3#, #x = 4# and #x = 5#.

Explanation:

#f(x) = x^3-12x^2+47x-60#

#color(white)()#
#underline("Discriminant")#

The discriminant #Delta# of a cubic #ax^3+bx^2+cx+d# is given by the formula:

#bb(Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd)#

In our example, #a=1#, #b=-12#, #c=47#, #d=-60# and we find:

#Delta = 318096-415292-414720-97200+609120 = 4 > 0#

Since #Delta > 0# we can deduce that #f(x)# has #3# distinct Real zeros.

The discriminant allows us to distinguish the cases:

  • #Delta > 0# - The cubic has #3# distinct Real zeros.
  • #Delta = 0# - The cubic has #3# Real zeros, at least two of which coincide.
  • #Delta < 0# - The cubic has #1# Real zero and a Complex conjugate pair of non-Real zeros.

#color(white)()#
#underline("Rational zeros")#

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 #60# and #q# a divisor of the coefficient #1# of the leading term.

Note that #f(-x) = -x^3-12x^2-47x-60# has coefficients which all have the same sign. So #f(x)# has no negative zeros.

Hence the only possible rational zeros are the positive factors of #60#, namely:

#1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60#

We could try each of these in turn to find our zeros, but let's try something different...

#color(white)()#
#underline("Tschirnhaus transformation")#

We can simplify the cubic, eliminating the square term by making a linear substitution - known as a Tschirnhaus transformation. In our example, we substitute #t=x-4# and find that the resulting cubic not only has no square term, it has no constant term either...

Note that:

#(x-4)^3 = x^3-12x^2+48x-64#

So:

#x^3-12x^2+47x-60=(x-4)^3-(x-4)#

Let #t=x-4#

Then

#f(x) = (x-4)^3-(x-4)#

#= t^3-t#

#= t(t^2-1)#

#= t(t^2-1^2)#

#= t(t-1)(t+1)#

#= (x-4)(x-5)(x-3)#

Hence the zeros of #f(x)# are #x=3#, #x=4# and #x=5#.