Question

Let $P$ be a non-zero polynomial such that $P(1+x)=P(1–x)$ for all real $x$, and $P\left(1\right)=0$. Let $m$ be the largest integer such that $(x–1{)}^m$ divides $P\left(x\right)$ for all such $P\left(x\right)$. Then $m$ equals

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$P\left(x\right)=0\Rightarrow (x-1)$ is a factor of $P\left(x\right)$.
$P(1+x)=P(1–x)$
Differentiating w.r.t. $x$
$P^{\prime }(1+x)=-P^{\prime }(1-x)$
Putting $x=0$
$P^{\prime }\left(1\right)=-P^{\prime }\left(1\right)\therefore P^{\prime }\left(1\right)=0$
Therefore, $(x-1)$ is a factor of $P^{\prime }\left(x\right)$
again differentiating and putting $x=0$
$P^{\prime \prime }(1+x)=P^{\prime \prime }(1-x)$
So we cannot deduce any thing about $P^{\prime \prime }\left(x\right)$
So the polynomial will be in form of
$P\left(x\right)=(x-1{)}^{2}Q(x)$
Hence, the value of $m=2$