Can we sketch the reciprocal of a polynomial?

Can we sketch the reciprocal of a polynomial?

Question

Express $x^3 - 3x - 2$ as the product of three linear factors.

Show that $x^3 - 3x - 2$ may also be expressed in the form $(x-1)^2(x+p) + q$, where $p$ and $q$ are constants to be found. Hence, or otherwise, show that the curve whose equation is $y = x^3 - 3x - 2$ has a minimum point where $x = 1$.

Sketch the curve whose equation is

$$\begin{equation*} y = \frac{1}{x^3 - 3x - 2}, \end{equation*}$$

indicating clearly all its asymptotes. Hence obtain the number of real roots of the equation

$$\begin{equation*} (x+1)(x^3 - 3x - 2) = 1. \end{equation*}$$

Solution

By the factor theorem, a polynomial $f(x)$ has a linear factor $(x-a)$ if and only if $f(a) = 0$.

If we let $f(x) = x^3 - 3x - 2$, we have

So $(x-2)$ is a factor of $f(x) = x^3 - 3x - 2$, and $f(x) = (x-2)(x^2+ax+1)$.

Since $f(x)$ has no $x^2$ term, $a = 2$, and $f(x)$ can be factorised completely as $f(x) = (x-2)(x+1)^2,$ which is the product of three linear factors.

By expanding and collecting powers of

$x$

, we see that

$$\begin{align*} (x-1)^2(x+p) + q &= (x^2 - 2x + 1)(x+p) + q \\ &= x^3 + px^2 - 2x^2 - 2px + x + p + q \\ &= x^3 + (p-2)x^2 + (1-2p)x + p + q. \end{align*}$$

Can we write for all

$x$

$$\begin{equation*} x^3 - 3x - 2 = x^3 + (p-2)x^2 + (1-2p)x + p + q \end{equation*}$$

for some $p$ and $q$?

By equating coefficients, we need

$p$

and

$q$

that satisfy

$$\begin{align*} 0 &= p - 2, \\ -3 &= 1 - 2p, \\ -2 &= p + q \end{align*}$$

simultaneously. The first gives us

$p = 2$

, as does the second, and the third becomes

$q = -4$

. So we can write down the identity

$$\begin{equation*} x^3 - 3x - 2 = (x-1)^2(x+2) - 4. \end{equation*}$$

We know $f(x) = (x-1)^2(x+2) - 4$. When $x = 1$, $f(x) = -4$.

When $x =1 + \alpha$, where $\alpha$ is small and positive, then $f(x) = \alpha^2(3+\alpha)-4 > -4$.

When $x = 1 - \alpha$, then $f(x) = \alpha^2(3-\alpha)-4 > -4$. Thus $y = f(x)$ has a minimum at $x = 1$.

Or else we can differentiate. If $y = f(x) = x^3 -3x -2$, then $y' = 3x^2-3 = 0$ when $x = 1$.

We have $y'' = 6x = 6$ when $x = 1$, which is positive, so we have a minimum.

We know

$$\begin{equation*} y = \frac{1}{x^3 - 3x - 2} = \frac{1}{(x-2)(x+1)^2} = \frac{1}{(x-1)^2(x+2) - 4}. \end{equation*}$$

In sketching the graph, we will ask four questions:

1. what are the intercepts?

When $x = 0$, $y = -\dfrac{1}{2}$. This is the $y$-intercept of the curve. There are no $x$-intercepts, since the numerator of $y$ can never equal zero.

1. what is the asymptotic behaviour?

There are two vertical asymptotes (corresponding to the roots of the denominator) at $x = -1$ and $x = 2$.

Since the denominator of $y$ tends towards $\pm \infty$ as $x$ tends towards $\pm \infty$, and as the numerator is constant, we can see $y \to 0$ as $x \to \pm \infty$. Thus there is one horizontal asymptote, $y = 0$.

1. where is $y$ positive and negative?

The sign of $y$ depends wholly on the sign of $(x-2)$, since $(x+1)^2$ is never negative, and so $y > 0$ if $x > 2$ and $y < 0$ if $x < 2$ (provided that $x \ne -1$).

1. are there any turning points?

Finally, the denominator has a local minimum at $x = 1$, and so its reciprocal $y$ must have a local maximum at $x = 1$.

We can now sketch $y$:

We can rewrite this equation as

$$\begin{equation*} x+1 = \frac{1}{x^3 - 3x - 2}. \end{equation*}$$

(Clearly $x^3 - 3x - 2$ cannot be zero, since in this case the equation is not satisfied.) A real solution to this equation corresponds precisely to an intersection between the line $y = x+1$ and the curve $y = \dfrac{1}{x^3 - 3x - 2}$.

Thus we have this sketch of the line and the curve. There are two intersections, and hence there are two real roots to the equation.