Review question

# How many real roots does $(x-1)(x+2)(x+3)=3x$ have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7792

## Solution

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

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.

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

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

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 $x = 1, -2$.

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

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:

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 $(x-1)(x+2)(x+3)=3x.$

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 $(x-1)(x+2)(x+3)=3x$.