The remainder obtained by dividing the polynomial $mx^2 - 3x + 6$ by $(x - 2)$ is twice the remainder obtained by dividing the polynomial $3x^2 + 5x - m$ by $(x + 3)$.
(a) Find the value of '$m$'.
(b) Using the value of '$m$', find the remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$.
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(a) By the Remainder Theorem, the remainder when a polynomial is divided by $(x − a)$ is the value of the polynomial at $x = a$.
Let R1 be the remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$. Put $x = 2$ :
$$\begin{aligned}\Rightarrow R_1 = m(2)^2 - 3(2) + 6 \\[1em] \Rightarrow R_1 = 4m - 6 + 6 \\[1em] \Rightarrow R_1 = 4m.\end{aligned}$$
Let R2 be the remainder when $3x^2 + 5x − m$ is divided by $(x + 3)$. Put $x = −3$ :
$$\begin{aligned}\Rightarrow R_2 = 3(-3)^2 + 5(-3) - m \\[1em] \Rightarrow R_2 = 27 - 15 - m \\[1em] \Rightarrow R_2 = 12 - m.\end{aligned}$$
According to the question, R1 = 2R2.
$$\begin{aligned}\Rightarrow 4m = 2(12 - m) \\[1em] \Rightarrow 4m = 24 - 2m \\[1em] \Rightarrow 4m + 2m = 24 \\[1em] \Rightarrow 6m = 24 \\[1em] \Rightarrow m = 4.\end{aligned}$$
Hence, the value of $\bm{m}$ = 4.
(b) The remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$ is R1 = $4m$.
$\Rightarrow R_1 = 4 \times 4 = 16.$
Hence, the remainder = 16.