Question

If the remainders of the polynomial $f(x)$ when divided by $x + 1,x - 2,x + 2$ are 6, 3, 15 then the remainder of $f(x)$ when divided by $(x + 1)(x + 2)(x - 2)$ is

• A. $2x^{2} - 3x + 1$
• B. $3x^{2} - 2x + 1$
• C. $2x^{2} - x - 3$
• D. $3x^{2} + 2x + 1$

Answer 1

Answer: (A)

$2x^{2} - 3x + 1$

Answer 2

$\frac{f(x)}{x + 1} = \varphi_{1}(x) + \frac{6}{x + 1},\frac{f(x)}{x - 2} = \varphi_{2}(x) + \frac{3}{x - 2}$ and

$\frac{f(x)}{x + 2} = \varphi_{3}(x) + \frac{15}{x + 2}$

$\frac{f(x)}{(x + 1)(x + 2)(x - 2)} = \varphi(x) + \frac{Q(x)}{(x + 1)(x + 2)(x - 2)}$

We have to find $Q(x)$, which will be a second degree polynomial. When $Q(x)$ is divided by $(x + 1)$, we should get the same remainder as being obtained by dividing $f(x)$ by $(x + 1)$ i.e., 6. Similarly when $Q(x)$ is divided by $(x - 2)$, remainder should be 3 and when $f(x)$ is divided by $x + 2,$ the remainder should be 15.

$\therefore$ $Q( - 1) = 6$

$Q(2) = 3$, $Q( - 2) = 15$

Let $Q(x) = \alpha x^{2} + \beta x + \gamma$, $\therefore$ $\alpha - \beta + \gamma = 6$ …..(i)

$4\alpha + 2\beta + \gamma = 3$ .....(ii); $4\alpha - 2\beta + \gamma = 15$ …..(iii)

$\Rightarrow \alpha = 2,\beta = - 3,\gamma = 1$; $\therefore Q(x) = 2x^{2} - 3x + 1$.