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P2 2024 Oct Q3

A Level / Edexcel / P2

IAL 2024 Oct Paper · Question 3

题目

Problem

f(x)=2x3−x2+Ax+Bf(x)=2x^3-x^2+Ax+B

where AA and BB are integers.

Given that when f(x)f(x) is divided by (x+3)(x+3) the remainder is 5555

(a) show that

3A−B=−1183A-B=-118
(2)

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

(b) find the value of AA and the value of BB.

(3)

(c) Hence find the quotient when f(x)f(x) is divided by (x−7)(x-7)

(2)
题目中文翻译 f(x)=2x3−x2+Ax+Bf(x)=2x^3-x^2+Ax+B

其中 AA、BB 为整数。已知 f(x)f(x) 除以 (x+3)(x+3) 的余数为 5555。

(a) 证明：(2)

3A−B=−1183A-B=-118

又已知 (2x−5)(2x-5) 是 f(x)f(x) 的因式。

(b) 求 AA、BB 的值。(3)

(c) 由此求 f(x)f(x) 除以 (x−7)(x-7) 所得的商。(2)

解答

(a)

解法一

思路

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

答题过程

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By the remainder theorem,

f(−3)= 552(−3)3−(−3)2+A(−3)+B= 55−54−9−3A+B= 55−3A+B= 1183A−B= −118.\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.

解法二

思路

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直接除以 x+3x+3，把余数与 5555 比较，得到与余式定理相同的条件。

答题过程

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Dividing f(x)f(x) by x+3x+3, the successive steps are

f(x)−(x+3)(2x2)= −7x2+Ax+B.\begin{align*} f(x)-(x+3)(2x^2)=&\,-7x^2+Ax+B. \end{align*}  −7x2+Ax+B−(x+3)(−7x)= (A+21)x+B.\begin{align*} &\,-7x^2+Ax+B-(x+3)(-7x)\\ =&\,(A+21)x+B. \end{align*}  (A+21)x+B−(x+3)(A+21)= B−3A−63.\begin{align*} &\,(A+21)x+B-(x+3)(A+21)\\ =&\,B-3A-63. \end{align*}

The remainder is 5555, so

B−3A−63= 553A−B= −118.(shown)\begin{align*} B-3A-63=&\,55\\ 3A-B=&\,-118.\quad\text{(shown)} \end{align*}

(b)

解法一

思路

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(2x−5)(2x-5) 是因式，所以 x=52x=\frac52 是根。代入 f(x)f(x) 得到第二个方程，再联立。

答题过程

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Since (2x−5)(2x-5) is a factor,

f(52)= 0.\begin{align*} f\big(\frac52\big)=&\,0. \end{align*}

So

2(52)3−(52)2+A(52)+B= 0.\begin{align*} 2\big(\frac52\big)^3-\big(\frac52\big)^2 +A\big(\frac52\big)+B=&\,0. \end{align*}

Multiply by 44:

125−25+10A+4B= 05A+2B= −50.\begin{align*} 125-25+10A+4B=&\,0\\ 5A+2B=&\,-50. \end{align*}

Now solve

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

From the first equation,

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

Substitute:

5A+2(3A+118)= −5011A+236= −5011A= −286A= −26.\begin{align*} 5A+2(3A+118)=&\,-50\\ 11A+236=&\,-50\\ 11A=&\,-286\\ A=&\,-26. \end{align*}

Then

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

Therefore,

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

解法二

思路

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除以已知因式 2x−52x-5，令余数为零，然后与 (a) 的条件联立。

答题过程

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Divide f(x)f(x) by 2x−52x-5:

f(x)−(2x−5)x2=4x2+Ax+B.f(x)-(2x-5)x^2=4x^2+Ax+B.  4x2+Ax+B−(2x−5)(2x)= (A+10)x+B.\begin{align*} &\,4x^2+Ax+B-(2x-5)(2x)\\ =&\,(A+10)x+B. \end{align*}  (A+10)x+B −(2x−5)(A2+5)= B+5A2+25.\begin{align*} &\,(A+10)x+B\\ &\,\hspace{2pt}-(2x-5)\bigg(\frac A2+5\bigg)\\ =&\,B+\frac{5A}{2}+25. \end{align*}

Since 2x−52x-5 is a factor, the remainder is zero:

5A+2B=−50.5A+2B=-50.

Together with 3A−B=−1183A-B=-118 from part (a),

6A−2B= −23611A= −286A= −26B= 3(−26)+118=40.\begin{align*} 6A-2B=&\,-236\\ 11A=&\,-286\\ A=&\,-26\\ B=&\,3(-26)+118=40. \end{align*}

(c)

解法一

思路

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先写出 f(x)=2x3−x2−26x+40f(x)=2x^3-x^2-26x+40，再除以 (x−7)(x-7)。题目问 quotient，不是问是否整除。

答题过程

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Using A=−26A=-26 and B=40B=40,

f(x)= 2x3−x2−26x+40.\begin{align*} f(x)=&\,2x^3-x^2-26x+40. \end{align*}

Divide by x−7x-7. The three subtraction steps are

f(x)−(x−7)(2x2)=13x2−26x+40.f(x)-(x-7)(2x^2)=13x^2-26x+40.  13x2−26x+40−(x−7)(13x)= 65x+40.\begin{align*} &\,13x^2-26x+40-(x-7)(13x)\\ =&\,65x+40. \end{align*} 65x+40−(x−7)(65)=495.65x+40-(x-7)(65)=495.

Hence

2x3−x2−26x+40= (x−7)(2x2+13x+65)+495.\begin{align*} 2x^3-x^2-26x+40 =&\,(x-7)(2x^2+13x+65)+495. \end{align*}

Therefore, the quotient is

2x2+13x+65.\begin{align*} 2x^2+13x+65. \end{align*}

也可以设商为 2x2+px+q2x^2+px+q、余数为常数 RR，比较

f(x)=(x−7)(2x2+px+q)+R.f(x)=(x-7)(2x^2+px+q)+R.

x2x^2 项给出 p−14=−1p-14=-1，所以 p=13p=13；xx 项给出 q−7p=−26q-7p=-26，所以 q=65q=65。这也得到相同的商，且不把 x−7x-7 错当作因式。