Review question

# How many real roots does this cubic polynomial have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8711

## Solution

Show that the equation $x^3 - 6x^2 + 12 = 0$ has two positive roots and one negative root.

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 $f(x)$.

$x$ $f(x)$
-2 -20
-1 5
0 12
1 7
2 -4
3 -15
4 -20
5 -13
6 12

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 $f(x)$.

$x$ $f(x)$
-1.5 -4.875
-1.4 -2.504
-1.3 -0.337
-1.2 1.632
-1.1 3.409
-1 5.0

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 $x = -1.3$.