2013-I-12

16-06-2021 app.cch No Comments

Ans: (a)

a = 3

,

b = - 1

and

c = 4

(b) No

Note that the remainder is $0$ . Then by factor theorem, we have
$\begin{array}{rcl} f (2) & = & 0 \\ 3 (2)^{3} - 7 (2)^{2} + k (2) - 8 & = & 0 \\ 2 k & = & 12 \\ k & = & 6 \end{array}$

By long division, we have

$\begin{aligned} 3 x^{2} - x \\ x - 2 & 3 x^{3} - 7 x^{2} + 6 x - 8 \\ \underset{―}{3 x^{3} - 6 x^{2}} \\ - x^{2} + 6 x \\ \underset{―}{- x^{2} + 2 x} \\ + 4 x - 8 \\ \underset{―}{+ 4 x - 8} \end{aligned}$

Hence, $f (x) \equiv (x - 2) (3 x^{2} - x + 4)$ .

Therefore, $a = 3$ , $b = - 1$ and $c = 4$ .
Consider the discriminant of $3 x^{2} - x + 4 = 0$ , we have
$\begin{array}{rcl} Δ & = & (- 1)^{2} - 4 (3) (4) \\ = & - 47 \\ < & 0 \end{array}$

Therefore, there is no real roots of the equation $3 x^{2} - x + 4 = 0$ .

Hence, there is only one real roots of $f (x) = 0$ . Therefore, I don’t agree.

2013, HKDSE-MATH, Paper 1 Polynomials

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