202110251042 Solution to 1967-CE-AMATH-1

When $14x^4-27x^3+ax^2+12x+9$ is divided by $(x-2)$ the remainder is $5$ . What is the value of $a$ ?

Solution.

$\begin{aligned} f(x) & \stackrel{\textrm{def}}{=} 14x^4-27x^3+ax^2+12x+9 \\ f(x) & = (x-2)g(x) + 5 \\ f(2) & = 5 \\ 5 & = 14(2)^4-27(2)^3+a(2)^2+12(2)+9 \\ a & = 9 \\ \end{aligned}$

$\therefore$ The value of $a$ is $9$ .

M	T	W	T	F	S	S
				1	2	3
4	5	6	7	8	9	10
11	12	13	14	15	16	17
18	19	20	21	22	23	24
25	26	27	28	29	30	31

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