Question

Let $r_1$ and $r_2$ be the remainders when the polynomials $p\left(x\right)=x^3+x^2-5kx-7$ and $q\left(x\right)=x^3+k{x}^2-12x+6$ are divided by $x+1$ and $x-2$ respectively. If $2r_1-r_2=10$, the value of $k$ is

• A. $\frac{7}{3}$
• B. $\frac{3}{7}$
• C. $\frac{5}{7}$
• D. $\frac{7}{5}$

Answer 1

The correct option is A

$\frac{7}{3}$

Explanation for correct answer:

Using remainder theorem

(Remainder theorem states that if a polynomial $f\left(x\right)$ is divided by $(x-a)$, then the remainder is given by $f\left(a\right)$.)

According to question, $p\left(x\right)=x^3+x^2-5kx-7$ is divided by $(x+1)$

$\therefore$ remainder $=$$p(-1)=r_1$

$$\begin{gathered} \Rightarrow {(-1)}^3+{(-1)}^2-5k(-1)-7=r_1 \\ \Rightarrow -1+1+5k-7=r_1 \\ \Rightarrow 5k-7=r_1\dots \dots \dots \left(1\right) \end{gathered}$$

Again, $q\left(x\right)=x^3+k{x}^2-12x+6$ is divided by $(x-2)$

$$\begin{gathered} q\left(2\right)=r_2 \\ \Rightarrow 2^3+k\times 2^2-12\times 2+6=r_2 \\ \Rightarrow 8+4k-24+6=r_2 \\ \Rightarrow 4k-10=r_2\dots \dots \dots \left(2\right) \end{gathered}$$

Given that $2r_1-r_2=10\dots \dots \dots \left(3\right)$

Finding the value of $k$

Using $\left(1\right)$ and $\left(2\right)$ in $\left(3\right)$ we get

$$\begin{gathered} 2(5k-7)-(4k-10)=10 \\ \Rightarrow 10k-14-4k+10=10 \\ \Rightarrow 6k-14=0 \\ \Rightarrow k=\frac{14}{6} \\ \therefore k=\frac{7}{3} \end{gathered}$$

Hence, the correct option is (A).