I came upon this question in my textbook, and it doesn't show you how they answered it, would someone be able to show me how to work it out (it was meant to be done using the factor theorem I believe):

When the polynomial $P(x)$ is divided by $x^2-1$ the remainder is $3x-1$. What is the remainder when $P(x)$ is divided by $x-1$?

Answers to this question will be very much appreciated! Thankyou


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We can write $$\begin{align}P(x)&=(x^2-1)Q(x)+3x-1\\&=(x-1)(x+1)Q(x)+3x-3+2\\&=(x-1)(x+1)Q(x)+3(x-1)+2\\&=(x-1)((x+1)Q(x)+3)+\color{red}{2}\end{align}$$


Hint Using factor theorem, $P(1)=3\cdot1-1=2$.


If $\frac{P(x)}{x^2-1}=Q(x)+\frac{3x-1}{x^2-1}$.

Then, multiplying both sides by $x^2-1$ gives $P(x)=(x^2-1) \cdot Q(x) + (3x-1)$.

We now divide $P(x)$ by (x-1), and obtain

$\frac{P(x)}{x-1}=\frac{(x^2-1) \cdot Q(x)+(3x-1)}{x-1}$.

To Simply the right side, note that $x^2-1=(x-1)(x+1)$. We can break apart the fraction across the addition in the numerator, and we can also rewrite the $-1$ at the end as $-3+2$ (you see why I do this in a moment).

This yields $\frac{P(x)}{x-1}=\frac{(x+1)(x-1) \cdot Q(x)}{x-1}+\frac{3x-3+2}{x-1}$.

Now, the division in the middle faction is easy, for the rightmost fraction, split again across the addition, and the factor out the $3$ in the numerator of the fraction with the $x$-term in it.

This gives $\frac{P(x)}{x-1}=(x+1)\cdot Q(x)+\frac{3(x-1)}{x-1}+\frac{2}{x-1}$.

Simplifying one last time, we obtain

$\frac{P(x)}{x-1}=(x+1)\cdot Q(x)+3+\frac{2}{x-1}$

and thus conclude that the remainder is $2$.

  • 2
    $\begingroup$ But $3x -1$ is the remainder not what it equals. So $\frac {P(x)}{x^2 -1} = Q(x) + \frac {3x-1}{x^2-1}$, where $Q(x)$ is the quotient $\endgroup$ – frog1944 Mar 12 '16 at 7:58
  • $\begingroup$ Good catch, I should read more carefully... $\endgroup$ – Justin Benfield Mar 12 '16 at 7:59
  • $\begingroup$ Corrected my answer now. $\endgroup$ – Justin Benfield Mar 12 '16 at 8:14

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