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CIE 9709 2024 March Paper 32 Q1

A Level / CIE / P3

CIE 9709 2024 March Paper 32 Paper · Question 1

题目

Problem

Find the quotient and remainder when x4−3x3+9x2−12x+27x^4-3x^3+9x^2-12x+27 is divided by x2+5x^2+5.

(3)
题目中文翻译

求 x4−3x3+9x2−12x+27x^4-3x^3+9x^2-12x+27 除以 x2+5x^2+5 所得的商和余式。

(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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按最高次项依次相除。每一步都减去“本次商项乘除式”,最后余式次数必须低于二次。

答题过程

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Let P(x)=x4−3x3+9x2−12x+27P(x)=x^4-3x^3+9x^2-12x+27. The first quotient term is x2x^2:

 P(x)−x2(x2+5)= −3x3+4x2−12x+27.\begin{align*} &\,P(x)-x^2(x^2+5)\\[4mm] =&\,-3x^3+4x^2-12x+27. \end{align*}

The next quotient term is −3x-3x:

 (−3x3+4x2−12x+27) −(−3x)(x2+5)= 4x2+3x+27.\begin{align*} &\,(-3x^3+4x^2-12x+27)\\[4mm] &\,\hspace{2pt}-(-3x)(x^2+5)\\[4mm] =&\,4x^2+3x+27. \end{align*}

The final quotient term is 44:

(4x2+3x+27)−4(x2+5)=3x+7.(4x^2+3x+27)-4(x^2+5)=3x+7.

Thus

P(x)=(x2+5)(x2−3x+4)+(3x+7).P(x)=(x^2+5)(x^2-3x+4)+(3x+7).

The quotient is x2−3x+4\boxed{x^2-3x+4} and the remainder is 3x+7\boxed{3x+7}.