# Where did I go wrong in applying the factor theorem?

Given that $$x + 1$$ and $$x - 3$$ are two of the four factors of the expression $$x^4 + px^3 + 5x^2 + 5x + q$$, find the values of $$p$$ and $$q$$.

I tried to answer this question using the factor theorem but got the answer wrong:

$$\text{Let } f(x) = x^4 + px^3 + 5x^2 + 5x + q$$

$$\text{Since } x + 1 \text{ and } x - 3 \text{ are factors of } f(x), \text{ then } f(-1) = 0 \text{ and } f(3) = 0, \text{ i.e.}$$

\begin{align} (-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \color{red}{\leftarrow (1)} \\ (3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \color{blue}{\leftarrow (2)} \end{align}

$$\text{From } \color{red}{(1)}:$$

\begin{align} (-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \\ 1 + p(-1) + 5(1) + (-5) + q &= 0 \\ 1 - p + 5 - 5 + q &= 0 \\ 1 - p + q &= 0 \\ q &= p - 1 \color{limegreen}{\leftarrow (3)} \end{align}

$$\text{From } \color{blue}{(2)}:$$

\begin{align} (3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \\ 81 + 27p + 45 + 15 + q &= 0 \\ 27p + q + 60 + 81 &= 0 \\ 27p + q + 141 &= 0 \\ q &= -27p - 144 \color{orange}{\leftarrow (4)} \end{align}

$$\color{orange}{(4)} + \color{limegreen}{(3)}:$$

\begin{align} -27p - 144 &= p - 1 \\ -27p - p &= 144 - 1 \\ -28p &= 143 \\ 28p &= -143 \\ p &= -\frac{143}{28} \\ \therefore p &= -5\frac{3}{28} \color{mediumpurple}{\leftarrow (5)} \end{align}

$$\text{Substitute } \color{mediumpurple}{(5)} \text{ into } \color{limegreen}{(3)}:$$

\begin{align} q &= -5\frac{3}{28} - 1 \\ \therefore q &= -6\frac{3}{28} \end{align}

The answers were $$p = -5, q = -6$$.

Where did I go wrong?

• You incorrectly wrote $-141$ as $-144$ in $(4)$. Commented Oct 19, 2020 at 12:10
• How did you go from 141 to 144 in consecutive lines? Commented Oct 19, 2020 at 12:13

$$q = -27p - \color{red}{141} {\leftarrow (4)}$$

Just another way.

Consider $$y=\frac{x^4 + px^3 + 5x^2 + 5x + q } {(x+1)(x-3) }$$ and perform the long division. You will have $$y=-\frac{q}{3}+\left(\frac{2 q}{9}-\frac{5}{3}\right) x-\left(\frac{7 q}{27}+\frac{5}{9}\right) x^2+\frac{1}{81} x^3 (-27 p+20 q-15)+\frac{1}{243} x^4 (54 p-61 q-96)+\cdots$$ So, $$-27 p+20 q-15=0$$ $$54 p-61 q-96=0$$

Solve for $$(p,q)$$ (simple).

Here is a way to see where you went wrong by comparison. Let \begin{align*} f(x) & = x^4+px^3 +5x^2+5x+q \\ g(x) & = (x+1)(x-3)(x-a)(x-b) \end{align*} Then write down the polynomial $$f-g=0$$. The coefficients must be all zero, i.e., we have $$a + b + p + 2=0,\; ab + 2a + 2b - 8=0,\; - 2ab + 3a + 3b - 5=0, q-3ab=0.$$ Solve for $$a$$ and $$b$$, and then put $$p=-a-b-2$$, $$q=3ab$$.

Your equation $$4$$ is not right.

It would be $$q=-27p-141$$

Then you will equate equation $$4$$ and $$3$$

$$-27p-141=p-1$$

You will get $$p$$ as $$-5$$.

Substitute $$p$$ as $$-5$$ in equation $$3$$ and we will get $$q$$ as $$-6$$