Review question

# If dividing by $x+a$ or $x-2a$ gives the same remainder, what's $a$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6488

## Solution

1. The expressions $px^4 - 5x + q$ and $x^4 - 2x^3 - px^2 - qx - 8$ have a common factor $x-2$. Find the value of $p$ and of $q$.

As $x-2$ is a factor of both of these polynomials, by the factor theorem, both expressions are zero when evaluated at $x = 2$ .

We thus have the simultaneous equations \begin{align*} p \times 2^4 - 5 \times 2 + q &= 0, \\ 2^4 - 2 \times 2^3 - p \times 2^2 - q \times 2 - 8 &= 0. \end{align*} That is, \begin{align*} 16p + q &= 10, \\ -4p - 2q &= 8. \end{align*}

The second equation becomes $q = -4-2p$ when dividing through by $2$. By substituting this into the first equation, we have that $16p + (-4-2p) = 10.$ That is, $14p = 14$ or $p = 1$. Consequently, $q = -4-2 = -6$.

1. The expression $x^3 - x^2 - 2x - 3$ has the same remainder when divided by $x+a$ and by $x-2a$. Find the non-zero values of $a$ and the corresponding values of the remainder.
The remainder theorem tells us that the expression evaluated at $x=-a$ is equal to the expression evaluated at $x=2a$. This gives us $\begin{equation} -a^3 - a^2 + 2a - 3 = 8a^3 - 4a^2 - 4a - 3. \label{eq:1} \end{equation}$

That is, $a$ is such that $9a^3 - 3a^2 - 6a = 0$ or $3a(3a^2 - a - 2) = 0.$ Factorising, we see that $3a^2 - a - 2 = (3a + 2)(a - 1)$ and so the non-zero values of $a$ that satisfy $\eqref{eq:1}$ are $a = 1$ and $a = -\dfrac{2}{3}$.

By substituting into either side of $\eqref{eq:1}$, the remainder in the case when $a=1$ is given by $-1^3 - 1^2 + 2 \times 1 - 3 = -3,$ and the remainder in the case when $a=-\dfrac{2}{3}$ is given by $-\left(-\frac{2}{3}\right)^3 - \left(-\frac{2}{3}\right)^2 + 2 \times \left( -\frac{2}{3} \right) - 3 = \frac{8 - 12 - 36 - 81}{27} = -\frac{121}{27}.$

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