2020-I-13 – Solving Master

Ans: (a)

- 8

(b) Yes

According to the division algorithm, let $f (x) = (x^{2} - 1) (a x + b) + (k x + 8)$ , where $a$ and $b$ are the real constants.
Since $f (x)$ is divisible by $x - 1$ , by the factor theorem, we have

$\begin{array}{rcl} f (1) & = & 0 \\ (1^{2} - 1) (a (1) + b) + k (1) + 8 & = & 0 \\ k & = & - 8 \end{array}$
Since $x + 3$ is a factor of $f (x)$ , by the factor theorem, we have
$\begin{array}{rcl} f (- 3) & = & 0 \\ [(- 3)^{2} - 1] [(a (- 3) + b] - 8 (- 3) + 8 & = & 0 \\ - 24 a + 8 b + 32 & = & 0 \dots ① \end{array}$

Since the remainder is $24$ when $f (x)$ is divided by $x$ , then by the remainder theorem, we have

$\begin{array}{rcl} f (0) & = & 24 \\ [(0)^{2} - 1] [a (0) + b] - 8 (0) + 8 & = & 24 \\ - b & = & 16 \\ b & = & - 16 \end{array}$

Sub. $b = - 16$ into $①$ , we have

$\begin{array}{rcl} - 24 a + 8 (- 16) + 32 & = & 0 \\ a & = & - 4 \end{array}$

Hence, we have

$\begin{array}{rcl} f (x) & = & 0 \\ (x^{2} - 1) (- 4 x - 16) - 8 x + 8 & = & 0 \\ (x + 1) (x - 1) (- 4 x - 16) - 8 (x - 1) & = & 0 \\ (x - 1) ([(x + 1) (- 4 x - 16) - 8] & = & 0 \\ (x - 1) (- 4 x^{2} - 20 x - 24) & = & 0 \\ - 4 (x - 1) (x^{2} + 5 x + 6) & = & 0 \\ - 4 (x - 1) (x + 2) (x + 3) & = & 0 \\ (x - 1) (x + 2) (x + 3) & = & 0 \end{array}$

Therefore, the roots are $1$ , $- 2$ and $- 3$ , which are all integers.

Hence, the claim is correct.