2012PP-I-10 – Solving Master

Ans: (a)

0

(b)

(x + 3) (2 x - 1) (3 x - 2)

It is given that $f (1) = 4$ . By the remainder theorem, the remainder of $f (x) \div (x - 1)$ is $4$ . By division algorithm, we have
$\begin{array}{rcl} f (x) & = & (x - 1) (6 x^{2} + 17 x - 2) + 4 \\ = & 6 x^{3} + 11 x^{2} - 19 x + 6 \\ f (- 3) & = & 6 (- 3)^{3} + 11 (- 3)^{2} - 19 (- 3) + 6 \\ = & 0 \end{array}$
Since $f (- 3) = 0$ , then by the factor theorem, $(x + 3)$ is a factor of $f (x)$ .
$\begin{aligned} 6 x^{2} - 7 x + 2 \\ x + 3 & 6 x^{3} + 11 x^{2} - 19 x + 6 \\ \underset{―}{6 x^{3} + 18 x^{2}} \\ - 7 x^{2} - 19 x \\ \underset{―}{- 7 x^{2} - 21 x} \\ + 2 x + 6 \\ \underset{―}{+ 2 x + 6} \end{aligned}$

Hence, we have

$\begin{array}{rcl} f (x) & = & 6 x^{3} + 11 x^{2} - 19 x + 6 \\ = & (x + 3) (6 x^{2} - 7 x + 2) \\ = & (x + 3) (2 x - 1) (3 x - 2) \end{array}$