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CIE 9709 2023 June Paper 33 Q2

A Level / CIE / P3

CIE 9709 2023 June Paper 33 Paper · Question 2

题目

Problem

Find the quotient and remainder when 2x4−272x^4-27 is divided by x2+x+3x^2+x+3.

(3)
题目中文翻译

求 2x4−272x^4-27 除以 x2+x+3x^2+x+3 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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依次用最高次项相除,并在每一步减去商项乘除式,直到余式次数低于二次。

答题过程

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The first quotient term is 2x22x^2:

(2x4−27)−2x2(x2+x+3)=−2x3−6x2−27.(2x^4-27)-2x^2(x^2+x+3) =-2x^3-6x^2-27.

The next term is −2x-2x:

(−2x3−6x2−27)−(−2x)(x2+x+3)=−4x2+6x−27.(-2x^3-6x^2-27)-(-2x)(x^2+x+3) =-4x^2+6x-27.

The final term is −4-4:

(−4x2+6x−27)−(−4)(x2+x+3)=10x−15.(-4x^2+6x-27)-(-4)(x^2+x+3)=10x-15.

Therefore

2x4−27= (x2+x+3)(2x2−2x−4) +(10x−15).\begin{align*} 2x^4-27=&\,(x^2+x+3)(2x^2-2x-4)\\[4mm] &\,\hspace{2pt}+(10x-15). \end{align*}

The quotient is 2x2−2x−4\boxed{2x^2-2x-4} and the remainder is 10x−15\boxed{10x-15}.

解法二

思路

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把被除式写为除式乘二次商再加一次余式,比较各次幂系数,依次求出五个常数。

答题过程

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Write

2x4−27= (x2+x+3)(Ax2+Bx+C) +Dx+E.\begin{align*} 2x^4-27=&\,(x^2+x+3)(Ax^2+Bx+C)\\[4mm] &\,\hspace{2pt}+Dx+E. \end{align*}

Comparing coefficients gives

A= 2,A+B= 0,3A+B+C= 0,3B+C+D= 0,3C+E= −27.\begin{align*} A=&\,2,\\[4mm] A+B=&\,0,\\[4mm] 3A+B+C=&\,0,\\[4mm] 3B+C+D=&\,0,\\[4mm] 3C+E=&\,-27. \end{align*}

Thus B=−2B=-2, C=−4C=-4, D=10D=10 and E=−15E=-15. The quotient is 2x2−2x−4\boxed{2x^2-2x-4} and the remainder is 10x−15\boxed{10x-15}.