CalcGospel 國際數學圖譜
返回

P2 2025 May A Q3

A Level / Edexcel / P2

IAL 2025 May A Paper · Question 3

题目

Problem

f(x)=(3x2−4x−5)(x−k)−5f(x)=(3x^2-4x-5)(x-k)-5

where kk is a constant.

(a) Deduce the value of the remainder when f(x)f(x) is divided by (x−k)(x-k)

(1)

Given that the remainder when f(x)f(x) is divided by (x+2)(x+2) is 2525

(b) show that the value of kk is −4-4

(2)

(c) Hence find the quotient and remainder when f(x)f(x) is divided by (3x−1)(3x-1)

(4)
题目中文翻译

f(x)=(3x2−4x−5)(x−k)−5f(x)=(3x^2-4x-5)(x-k)-5，其中 kk 是常数。

(a) 推出 f(x)f(x) 除以 (x−k)(x-k) 时的余数。

(1)

已知 f(x)f(x) 除以 (x+2)(x+2) 时的余数为 2525。

(b) 证明 kk 的值为 −4-4。

(2)

(c) 由此求 f(x)f(x) 除以 (3x−1)(3x-1) 时的商和余数。

(4)

解答

(a)

解法一

思路

展开

除以 x−kx-k 的余数是 f(k)f(k)。代入后前面乘积为 00。

答题过程

展开

By the remainder theorem, the remainder is f(k)f(k).

f(k)= (3k2−4k−5)(k−k)−5= −5.\begin{align*} f(k)=&\,(3k^2-4k-5)(k-k)-5\\ =&\,-5. \end{align*}

Therefore, the remainder is

−5.\begin{align*} -5. \end{align*}

(b)

解法一

思路

展开

除以 x+2x+2 的余数是 f(−2)f(-2)。题目说余数为 2525，所以令 f(−2)=25f(-2)=25。

答题过程

展开

By the remainder theorem,

f(−2)=25.\begin{align*} f(-2)=25. \end{align*}

So

3(−2)2−4(−2)−5= 12+8−5= 15.\begin{align*} 3(-2)^2-4(-2)-5=&\,12+8-5\\ =&\,15. \end{align*}

Therefore,

15(−2−k)−5= 2515(−2−k)= 30−2−k= 2k= −4.\begin{align*} 15(-2-k)-5=&\,25\\ 15(-2-k)=&\,30\\ -2-k=&\,2\\ k=&\,-4. \end{align*}

This is the required result.

(c)

解法一

思路

展开

先把 k=−4k=-4 代入并展开，再对 3x−13x-1 做多项式除法。

答题过程

展开

Since k=−4k=-4,

f(x)= (3x2−4x−5)(x+4)−5= 3x3+12x2−4x2 −16x−5x−20−5= 3x3+8x2−21x−25.\begin{align*} f(x)=&\,(3x^2-4x-5)(x+4)-5\\ =&\,3x^3+12x^2-4x^2\\ &\,\hspace{2pt}-16x-5x-20-5\\ =&\,3x^3+8x^2-21x-25. \end{align*}

Now divide by 3x−13x-1. The first quotient term is x2x^2:

 3x3+8x2−21x−25 −(3x3−x2)= 9x2−21x−25.\begin{align*} &\,3x^3+8x^2-21x-25\\ &\,\hspace{2pt}-(3x^3-x^2)\\ =&\,9x^2-21x-25. \end{align*}

The next quotient term is 3x3x:

 9x2−21x−25−(9x2−3x)= −18x−25.\begin{align*} &\,9x^2-21x-25-(9x^2-3x)\\ =&\,-18x-25. \end{align*}

The final quotient term is −6-6:

−18x−25−(−18x+6)=−31.-18x-25-(-18x+6)=-31.

Therefore, the quotient is

x2+3x−6\begin{align*} x^2+3x-6 \end{align*}

and the remainder is

−31.\begin{align*} -31. \end{align*}

解法二

思路

展开

也可以设

f(x)=(3x−1)(Ax2+Bx+C)+Rf(x)=(3x-1)(Ax^2+Bx+C)+R

然后比较系数。这样不需要写长除法。

答题过程

展开

From the expansion above,

f(x)=3x3+8x2−21x−25.\begin{align*} f(x)=3x^3+8x^2-21x-25. \end{align*}

Let

f(x)= 3x3+8x2−21x−25,f(x)= (3x−1)(Ax2+Bx+C)+R.\begin{align*} f(x)=&\,3x^3+8x^2-21x-25,\\ f(x)=&\,(3x-1)(Ax^2+Bx+C)+R. \end{align*}

Expand the right hand side:

f(x)= 3Ax3+(3B−A)x2 +(3C−B)x−C+R.\begin{align*} f(x)=&\,3Ax^3+(3B-A)x^2\\ &\,\hspace{2pt}+(3C-B)x-C+R. \end{align*}

Compare coefficients:

3A= 3A= 1,3B−A= 83B−1= 8B= 3,3C−B= −213C−3= −21C= −6,−C+R= −256+R= −25R= −31.\begin{align*} 3A=&\,3\\ A=&\,1,\\[2mm] 3B-A=&\,8\\ 3B-1=&\,8\\ B=&\,3,\\[2mm] 3C-B=&\,-21\\ 3C-3=&\,-21\\ C=&\,-6,\\[2mm] -C+R=&\,-25\\ 6+R=&\,-25\\ R=&\,-31. \end{align*}

Therefore, the quotient is

x2+3x−6\begin{align*} x^2+3x-6 \end{align*}

and the remainder is

−31.\begin{align*} -31. \end{align*}

As a check, the remainder theorem gives

R= f(13)= 19+89−7−25= −31.\begin{align*} R=&\,f\bigg(\frac13\bigg)\\ =&\,\frac19+\frac89-7-25\\ =&\,-31. \end{align*}

The quotient above is still required; finding only the remainder does not complete part (c).