How many real roots does this cubic polynomial have?

How many real roots does this cubic polynomial have?

Question

Show that the equation $x^3 - 6x^2 + 12 = 0$ has two positive roots and one negative root. Show that the negative root lies between $-1$ and $-1.5$ and find it correct to one decimal place.

Suggestion

If we cannot deduce directly how to factorise a polynomial, do we know a way of honing in on the location of (any) roots?

Solution

Let $f(x) = x^3 - 6x^2 + 12$, which is a cubic, and thus has at most three real roots. We can tabulate various values for $f(x)$.

The function $f(x)$ changes sign three times in the table, so that $f(x)$ has three real roots, one between $x = -2$ and $x = -1$; one between $x = 1$ and $x = 2$; and a third between $x = 5$ and $x = 6$.

So $f(x)$ has two positive roots and one negative root.

Show that the negative root lies between $-1$ and $-1.5$…

We saw above that $f(-1) = 5$. As $f(-1.5) = -\dfrac{39}{8} < 0$, the negative root must lie between $x = -1$ and $x = -1.5$.

…and find it correct to one decimal place.

We can tabulate more values for $f(x)$.

Thus the negative root lies between $x = -1.3$ and $x = -1.2$.

We have $f(-1.25) = 0.671875 > 0$, and so the root falls between $x = -1.3$ and $x = -1.25$, so that, to one decimal place, the negative root is $x = -1.3$.