题目

Problem

$f(x)=ax^3+3x^2-8x+2 \quad$ where $a$ is a constant

Given that when $f(x)$ is divided by $(x-2)$ the remainder is $3$ , find the value of $a$ .

(3)

解答

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除以 $x-2$ 的余数就是 $f(2)$ 。题目说余数是 $3$ ，所以直接令 $f(2)=3$ 。

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By the remainder theorem,

\begin{align*} f(2)=&\,3\\ a(2)^3+3(2)^2-8(2)+2=&\,3\\ 8a+12-16+2=&\,3\\ 8a-2=&\,3\\ 8a=&\,5\\ a=&\,\frac58. \end{align*}

Therefore,

\begin{align*} a=\frac58. \end{align*}

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也可以把余数条件改写成： $f(x)-3$ 被 $x-2$ 整除。这样 $x=2$ 是 $f(x)-3$ 的根，代入后同样得到 $a$ 。

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Since the remainder is $3$ , the expression $f(x)-3$ is divisible by $(x-2)$ .

So $x=2$ is a root of $f(x)-3$ :

\begin{align*} f(2)-3 =&\,\, 0\\[4mm] \left(a(2)^3+3(2)^2-8(2)+2\right)-3 =&\,\, 0\\[4mm] 8a+12-16+2-3 =&\,\, 0\\[4mm] 8a-5 =&\,\, 0\\[4mm] a =&\,\, \frac58. \end{align*}

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通过待定系数法（inspection）。如果 $f(x)-3$ 能被 $x-2$ 整除，可设商式为二次多项式 $ax^2+px+q$ 。对比常数项可以快速求出 $q$ ，再通过一次项与二次项系数求出 $a$ 。

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Since $f(x)-3$ is divisible by $(x-2)$ , we can write:

\begin{align*} f(x)-3 = (x-2)(ax^2+px+q) \end{align*}

Substitute $f(x) = ax^3+3x^2-8x+2$ :

\begin{align*} ax^3+3x^2-8x-1 = ax^3+(p-2a)x^2+(q-2p)x-2q \end{align*}

Comparing the constant terms on both sides:

\begin{align*} -2q = -1 \implies q = \frac12 \end{align*}

Comparing the coefficients of $x^2$ :

\begin{align*} p-2a = 3 \implies p = 2a+3 \end{align*}

Comparing the coefficients of $x$ :

\begin{align*} q-2p =&-8\\[4mm] \frac12 - 2(2a+3) =&-8\\[4mm] \frac12 - 4a - 6 =&-8\\[4mm] -4a =&-2.5\\[4mm] a =&\,\, \frac58. \end{align*}

Therefore, the value of $a$ is $\frac{5}{8}$ .