Question

If polynomials $a{x}^3+3x^2-3$ and $2x^3-5x+a$ when divided by $(x-4)$ leave the remainders as $R_1$ and $R_2$ respectively.

Find the values of $a$ in $2R_1-R_2=0$

Answer 1

Find the value of $a$

If $x-m$ is the quotient and $a_1{x}^n+a_2{x}^{n-1}+a_3{x}^{n-2}....+a_{n+1}{x}^0$ is the dividend, then upon substituting the value of $m$ on the dividend in place of $x$ we get the remainder.

Let the remainder be $R_1$ and $R_2$ as given :

$$\begin{gathered} R_1=a{x}^3+3x^2-3 \\ =a{\left(4\right)}^3+3{\left(4\right)}^2-3=64a+45 \end{gathered}$$

$$\begin{gathered} R_2=2x^3-5x+a \\ =2{\left(4\right)}^3-5\left(4\right)+a \\ =128-20+a=108+a \end{gathered}$$

Given: $2R_1-R_2=0$

Now put the value of $R_1$and $R_2$ in the given equation

$\therefore 2(64a+45)-(108+a)=0$

$$\begin{gathered} \Rightarrow 128a+90-108-a=0 \\ \Rightarrow 127a=18 \\ \Rightarrow a=\frac{18}{127} \end{gathered}$$

Hence the value of $a$ is $\frac{18}{127}$