题目

Problem

$f(x)=2x^3-x^2+Ax+B$

where $A$ and $B$ are integers.

Given that when $f(x)$ is divided by $(x+3)$ the remainder is $55$

(a) show that

3A-B=-118

(2)

Given also that $(2x-5)$ is a factor of $f(x)$ ,

(b) find the value of $A$ and the value of $B$ .

(3)

(2)

解答

(a)

解法一

思路

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除以 $x+3$ 的余数是 $f(-3)$ ，所以 $f(-3)=55$ 。

答题过程

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By the remainder theorem,

\begin{align*} f(-3)=&\,55\\ 2(-3)^3-(-3)^2+A(-3)+B=&\,55\\ -54-9-3A+B=&\,55\\ -3A+B=&\,118\\ 3A-B=&\,-118. \end{align*}

This is the required result.

(b)

解法一

思路

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$(2x-5)$ 是因式，所以 $x=\frac52$ 是根。代入 $f(x)$ 得到第二个方程，再联立。

答题过程

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Since $(2x-5)$ is a factor,

\begin{align*} f\left(\frac52\right)=0. \end{align*}

\begin{align*} 2\left(\frac52\right)^3-\left(\frac52\right)^2 +A\left(\frac52\right)+B=&\,0. \end{align*}

Multiply by $4$ :

\begin{align*} 125-25+10A+4B=&\,0\\ 5A+2B=&\,-50. \end{align*}

Now solve

\begin{align*} 3A-B=&\,-118,\\ 5A+2B=&\,-50. \end{align*}

From the first equation,

\begin{align*} B=3A+118. \end{align*}

Substitute:

\begin{align*} 5A+2(3A+118)=&\,-50\\ 11A+236=&\,-50\\ 11A=&\,-286\\ A=&\,-26. \end{align*}

Then

\begin{align*} B=3(-26)+118=40. \end{align*}

Therefore,

\begin{align*} A=-26,\qquad B=40. \end{align*}

(c)

解法一

思路

展开

先写出 $f(x)=2x^3-x^2-26x+40$ ，再除以 $(x-7)$ 。题目问 quotient，不是问是否整除。

答题过程

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Using $A=-26$ and $B=40$ ,

\begin{align*} f(x)=2x^3-x^2-26x+40. \end{align*}

Divide by $(x-7)$ :

\begin{align*} 2x^3-x^2-26x+40 =&\,(x-7)(2x^2+13x+65)+495. \end{align*}

Therefore, the quotient is

\begin{align*} 2x^2+13x+65. \end{align*}

IAL 2024 Oct Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程