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Polynomial $p(x)$ leaves a remainder of $4$ when divided by $x-1$ and a remainder of $-2$ when divided by $x+1$.

Find the remainder when $p(x)$ is divided by $x^2 -1$ .

According to Remainder Theorem, when a polynomial $p(x)$ is divided by $(ax+b)$, the remainder is $p\left( -\frac {b}{a}\right)$ .

So, I did the following:

\begin{align}p(1)&=4\\ p(-1)&= -2\end{align}

\begin{align}p(x)&= (x^2-1)q(x) + Ax+ B\\ p(x)&= (x-1)(x+1)q(x) + Ax+B\end{align}

When \begin{align}p(1) &= A(1) + B\\ A+B&=4\tag{1}\end{align}

When \begin{align}p(-1)&= -A+ B\\ -A+B&= -2\tag{2}\end{align}

Doing $(1)+(2)$ gives:

\begin{align}2B&=2\\ B&=1\end{align}

Substitute $B=1$ into $(1)$ gives $A=3$

So, I got the remainder as $3x+ 1$

But, the answer in the book is $x+3$ , which means my values of $A$ and $B$ have been mixed up.

Please tell me where I went wrong

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  • $\begingroup$ If we divide $x+3$ by $x+1$ , the remainder is clearly $2$. Weird ! $\endgroup$
    – Peter
    Jan 23 '20 at 11:00
  • $\begingroup$ @Peter Why would you divide the remainder by the factor? $\endgroup$
    – gc3941d
    Jan 23 '20 at 11:05
  • $\begingroup$ Apparently, I mixed the answer (which gives the remainder) with the polyomial, sorry. $\endgroup$
    – Peter
    Jan 23 '20 at 11:13
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$3x+1 = 3(x-1) + 4\\3x+1 = 3(x+1)-2$

So your answer $3x+1$ is correct.

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  • $\begingroup$ I see, thank you $\endgroup$
    – gc3941d
    Jan 23 '20 at 12:04
  • $\begingroup$ What is missing : The proof that this is the only polynomial modulo $x^2-1$ that does the job. Otherwise we could not conclude the remainder. $\endgroup$
    – Peter
    Jan 23 '20 at 12:09
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    $\begingroup$ @Peter Remainder is unique because if there are two remainders $r$ and $r'$ then their difference must be divisible by $x^2-1$. But since the degrees of both $r$ and $r'$ are less than the degree of $x^2-1$, we must have $r-r'=0$, so $r=r'$. $\endgroup$
    – gandalf61
    Jan 23 '20 at 12:41

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