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CIE 9709 2025 November Paper 33 Q2

A Level / CIE / P3

CIE 9709 2025 Nov Paper 33 Paper · Question 2

题目

Problem

Find the quotient and the remainder when 3x4−2x23x^4-2x^2 is divided by x+1x+1.

(3)
题目中文翻译

求 3x4−2x23x^4-2x^2 除以 x+1x+1 所得的商和余数。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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补上缺失项的零系数后做综合除法,除数 x+1x+1 对应使用 −1-1。

答题过程

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Dividing by x+1x+1, use the synthetic value −1-1 and coefficients 3,0,−2,0,03,0,-2,0,0:

StepLeading coefficientNext coefficient calculation
1330−3=−30-3=-3
2−3-3−2+3=1-2+3=1
3110−1=−10-1=-1
4−1-10+1=10+1=1

Thus

3x4−2x2= (x+1)(3x3−3x2+x−1) +1.\begin{align*} 3x^4-2x^2=&\,(x+1)(3x^3-3x^2+x-1)\\[4mm] &\,\hspace{2pt}+1. \end{align*}

The quotient is 3x3−3x2+x−1\boxed{3x^3-3x^2+x-1} and the remainder is 1\boxed{1}.

解法二

思路

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先用余式定理求常数余式,再对商设一般形式并比较系数。

答题过程

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Let f(x)=3x4−2x2f(x)=3x^4-2x^2. By the remainder theorem, the remainder is

R=f(−1)=3−2=1.R=f(-1)=3-2=1.

Write

3x4−2x2= (x+1)(Ax3+Bx2+Cx+D) +1.\begin{align*} 3x^4-2x^2=&\,(x+1)(Ax^3+Bx^2+Cx+D)\\[4mm] &\,\hspace{2pt}+1. \end{align*}

Comparing coefficients gives

A= 3,A+B= 0,B+C= −2,C+D= 0.\begin{align*} A=&\,3,\\[4mm] A+B=&\,0,\\[4mm] B+C=&\,-2,\\[4mm] C+D=&\,0. \end{align*}

Therefore A=3A=3, B=−3B=-3, C=1C=1, D=−1D=-1. The constant coefficient checks since D+1=0D+1=0.

Hence the quotient is 3x3−3x2+x−1\boxed{3x^3-3x^2+x-1} and the remainder is 1\boxed{1}.