Question

If the expressions $a{x}^3+3x^2-3$ and $2x^3-5x+a$ on dividing by $x-4$ leave the same remainder, then the value of $a$ is :

• A. $1$
• B. $0$
• C. $2$
• D. $-1$

Answer 1

Answer: (A)

Remainder theorem: If $P\left(x\right)$ is a polynomial and it is divided by another polynomial $(x-a)$ then the remainder is $P\left(a\right)$.

Let, $f_1\left(x\right)=a{x}^3+3x^2-3$,

If $f_1\left(x\right)$ is divided by $x-4$ then the remainder is $f_1\left(4\right)$

$\Rightarrow f_1\left(4\right)=a\times \left(4^3\right)+3\times \left(4^2\right)-3$

$\Rightarrow f_1\left(4\right)=64a+48-3$

$\Rightarrow f_1\left(4\right)=64a+45$---(1)

Let, $f_2\left(x\right)=2x^3-5x+a$

If $f_2\left(x\right)$ is divided by $x-4$ then the remainder is $f_2\left(4\right)$.

$\Rightarrow f_2\left(4\right)=2\times \left(4^3\right)-5\times \left(4\right)+a$

$\Rightarrow f_2\left(4\right)=128-20+a$

$\Rightarrow f_2\left(4\right)=108+a$ ---(2)

It is given that, $f_1\left(x\right)$ and $f_2\left(x\right)$ are divided by $(x-4)$ leaves the same remainder.

$\Rightarrow f_1\left(4\right)=f_2\left(4\right)$

$\Rightarrow 64a+45=108+a$

$\Rightarrow 64a-a=108-45$

$\Rightarrow 63a=63$

$\Rightarrow a=\frac{63}{63}$

$\Rightarrow a=1$

Hence, Option B is correct.