Question

When a polynomial $P\left(x\right)$ is divided by $x,(x-2)$ and $(x-3)$, remainders are $1$, $3$ and $2$ respectively. the same polynomial is divided by $x(x-2)(x-3)$, the remainder is $a{x}^2+bx+c$, then the value of $c$ is

• A. $-3$
• B. $-2$
• C. $6$
• D. $1$

Answer 1

Answer: (D)

$1$

According to factor theorem

$p\left(x\right)=xq\left(x\right)+1$

$P\left(x\right)=(x-2)q^{\prime }\left(x\right)+3$

$P\left(x\right)=(x-3)q^{\prime \prime }\left(x\right)+2$

So, At $x=0,P\left(0\right)=1$

At $x=2,P\left(2\right)=3$

At $x=3,P\left(3\right)=2$

Now when the polynomial $P\left(x\right)$ is divided by $(x-2)(x-3)x$ the remainder must have the degree less than $3$ . that is the remainder will be of the form $a{x}^2+bx+c$

$⟹P\left(x\right)=x(x-2)(x-3)q^{\prime \prime \prime \prime }\left(x\right)+a{x}^2+bx+c$

So, $P\left(0\right)=1=a(0{)}^2+b(0)+c⟹c=1$