题目
Problem
Find the quotient and remainder when 2x4−27 is divided by x2+x+3.
(3)
题目中文翻译
求 2x4−27 除以 x2+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 2x2:
(2x4−27)−2x2(x2+x+3)=−2x3−6x2−27.
The next term is −2x:
(−2x3−6x2−27)−(−2x)(x2+x+3)=−4x2+6x−27.
The final term is −4:
(−4x2+6x−27)−(−4)(x2+x+3)=10x−15.
Therefore
2x4−27=(x2+x+3)(2x2−2x−4)+(10x−15).
The quotient is 2x2−2x−4 and the remainder is 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.
Comparing coefficients gives
A=A+B=3A+B+C=3B+C+D=3C+E=2,0,0,0,−27.
Thus B=−2, C=−4, D=10 and E=−15.
The quotient is 2x2−2x−4 and the remainder is 10x−15.