How many real roots does $(x-1)(x+2)(x+3)=3x$ have?

How many real roots does $(x-1)(x+2)(x+3)=3x$ have?

Question

Express $\dfrac{3x}{(x-1)(x+2)}$ in partial fractions.

Show that $\dfrac{dy}{dx}$ is negative at all points on the graph of $y=\frac{3x}{(x-1)(x+2)}.$

Sketch this graph, showing the two asymptotes parallel to the $y$-axis and the asymptote perpendicular to the $y$-axis.

By sketching on the same diagram a second graph (the equation of which should be stated), or otherwise, find the number of real roots of the equation $(x-1)(x+2)(x+3)=3x.$

Suggestion

How does using partial fractions help us to differentiate the expression?

Things to look out for when curve sketching:

• asymptotes;

• intercepts;

• stationary points and their nature;

• behaviour as $x\rightarrow \pm \infty$.

Can we make this look something like the graph we’ve just sketched?

Could we make the roots of the equation into the intersections between the original graph and the new graph?

Solution

Let $\frac{3x}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$

Then by multiplying through by $(x-1)(x+2)$, we have $3x = A(x+2) + B(x-1).$

If we let

$x = 1$

, then our equation becomes

$$\begin{align*} 3 = 3A, \\ \text{so} \qquad A = 1. \end{align*}$$

If we now choose to use

$x = -2$

, then we have

$$\begin{align*} -6 = -3B, \\ \text{so} \qquad B = 2. \end{align*}$$

Alternatively, we can compare coefficients.

Comparing coefficients

We first need to expand the terms on the right hand side of the equation:

$$\begin{align*} 3x = Ax + 2A + Bx - 1B, \\ \text{so} \qquad 3x = (A + B)x + (2A - B). \end{align*}$$

Looking at the constant terms on each side of the equation, we see that $2A - B = 0$, and looking at the terms in $x$ on each side, we find $3 = A + B$.

Adding these two equations gives $A = 1$, and from there $B = 2$.

So we have $\frac{3x}{(x-1)(x+2)} = \frac{1}{x-1} + \frac{2}{x+2}.$

We differentiate the partial fractions to obtain $\frac{dy}{dx} = -\frac{1}{(x-1)^2} - \frac{2}{(x+2)^2}.$

We can see that this must be negative at all points, because each term is negative.

The gradient is not defined at $x = 1, -2$.

When $x = 0$, $y = 0$. There are no other points where the curve crosses the axes.

The two asymptotes parallel to the $y$-axis are $x=1$ and $x=-2$.

We have determined that $\dfrac{dy}{dx}$ is always negative, and so it cannot equal $0$, and thus there are no stationary points.

We also observe that as $x\rightarrow \pm \infty$, $y\rightarrow 0$.

We can now confidently sketch the graph:

We observe that the equation can be rearranged to $x+3=\frac{3x}{(x-1)(x+2)},$ which suggests sketching the graph $y=x+3$ onto the original graph.

There are three points of intersection for the two graphs, so there are three real roots for the equation $(x-1)(x+2)(x+3)=3x$.