If this cubic has two equal roots, what can $p$ be?

If this cubic has two equal roots, what can $p$ be?

Question

A polynomial $f(x)$ has a factor $(x-a)^2$. Show that $(x-a)$ is a factor of $f'(x)$.

The equation $x^3 - 4x^2 - 3x + p = 0$ has two equal roots; find the two possible values of $p$.

Solution

If

$f(x)$

has a factor

$(x-a)^2$

then by we can write

$$\begin{equation*} f(x) = (x-a)^2g(x) \end{equation*}$$

for a polynomial

$g(x)$

(which may be constant). Thus, by the

product rule

$$\begin{align*} f'(x) &= \frac{d}{dx} \left( (x-a)^2 \right) g(x) + (x-a)^2 \frac{d}{dx} \left( g(x) \right) \\ &= 2(x-a)g(x) + (x-a)^2g'(x) \\ &= (x-a) \left( 2g(x) + (x-a)g'(x) \right). \end{align*}$$

Since $g(x)$ and $g'(x)$ are polynomials, we have shown that $f'(x)$ has $(x-a)$ as a factor.

We can understand this result graphically as follows. If $(x-a)^2$ is a factor, then $f(x)=0$ has a repeated root at $x=a$ and the graph of $y=f(x)$ touches the $x$-axis here. The gradient of the graph must therefore be zero at this point, $f'(a)=0$ and therefore $(x-a)$ is a factor of $f'(x)$.

Let $f(x) = x^3 - 4x^2 - 3x + p$, and let’s assume the repeated root is at $x = a$. Then $(x-a)^2$ is a factor of $f(x)$.

From the result above,

$(x-a)$

is a factor of

$$\begin{equation*} f'(x) = 3x^2 - 8x - 3 = (3x + 1)(x - 3) \end{equation*}$$

so that $a = 3$ or $a = -\dfrac{1}{3}$.

If $a = 3$, then substituting $x=a$ into $x^3-4x^2-3x+p=0$, we have $p = 18$.

If $a = -\dfrac{1}{3}$, then from $x^3-4x^2-3x+p=0,$ we have $p = -\dfrac{14}{27}$.

Thus $p = 18$ and $p = -\dfrac{14}{27}$ are the two possible values for $p$.