Question

Let $p\left(x\right)$ be a polynomial, which when divided by $x-3$ and $x-5$ leaves remainders $10$ and $6$, respectively. If the polynomial is divided by $(x-3)(x-5)$, then the remainder is :

• A. $-2x+16$
• B. $16$
• C. $2x-16$
• D. $60$

Answer 1

The correct option is A $-2x+16$

Given, $p\left(x\right)$ when divided by $x-3$ and $x-5$, leaves remainder $10$ and $6$, respectively.

From remainder theorem of polynomial, $p\left(3\right)=10$ and $P\left(5\right)=6$.

If the polynomial is divided by $(x-3)(x-5)$, then the remainder must be of the form $ax+b$ (Since, degree of remainder is less than that of the divisor).

$\Rightarrow p\left(x\right)=q\left(x\right)(x-3)(x-5)+(ax+b)$, where $q\left(x\right)$ is some polynomial (quotient obtained).

Substituting for $x=3$ and $x=5$:

$p\left(3\right)=q\left(x\right)(3-3)(3-5)+a\left(3\right)+b=0+3a+b=3a+b⟹10=3a+b--\left(i\right)$

$p\left(5\right)=q\left(x\right)(5-3)(5-5)+a\left(5\right)+b=0+5a+b=5a+b⟹6=5a+b--\left(ii\right)$.

Solving for $a$ and $b$, we get,

$\left(i\right)-\left(ii\right)⟹10-6=(3-5)a⟹4=-2a⟹a=-2--\left(a\right)$.

$\left(a\right)$ in $\left(ii\right)⟹6=5(-2)+b⟹6=-10+b⟹b=6+10⟹b=16$.

$\therefore a=-2$ and $b=16$.

$\Rightarrow$ Remainder $=-2x+16$.

Therefore, option $A$ is correct.