Question

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

Answer 1

`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`.