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

##### 1 Answer

#### Answer:

#### Explanation:

The discriminant

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

In our example,

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

Since

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.

By the rational root theorem, any rational zeros of

Note that

Hence the only possible rational zeros are the positive factors of

#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...

We can simplify the cubic, eliminating the square term by making a linear substitution - known as a Tschirnhaus transformation. In our example, we substitute

Note that:

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

So:

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

Let

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