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IAL 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)

解答

(a)

解法一

思路

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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

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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

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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

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By the remainder theorem,

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

So

(3(−2)2−4(−2)−5)(−2−k)−5= 25(12+8−5)(−2−k)−5= 2515(−2−k)−5= 2515(−2−k)= 30−2−k= 2k= −4.\begin{align*} (3(-2)^2-4(-2)-5)(-2-k)-5=&\,25\\ (12+8-5)(-2-k)-5=&\,25\\ 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

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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

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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^2-16x-5x-20-5\\ =&\,3x^3+8x^2-21x-25. \end{align*}

Now divide by 3x−13x-1:

3x3+8x2−21x−25= (3x−1)(x2+3x−6)−31.\begin{align*} 3x^3+8x^2-21x-25 =&\,(3x-1)(x^2+3x-6)-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*}

解法二

思路

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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

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From the expansion above,

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

Let

3x3+8x2−21x−25=(3x−1)(Ax2+Bx+C)+R.3x^3+8x^2-21x-25 =(3x-1)(Ax^2+Bx+C)+R.

Expand the right hand side:

(3x−1)(Ax2+Bx+C)+R= 3Ax3+(3B−A)x2 +(3C−B)x+(−C+R).\begin{align*} (3x-1)(Ax^2+Bx+C)+R =&\,3Ax^3+(3B-A)x^2\\ &\,\hspace{2pt}+(3C-B)x+(-C+R). \end{align*}

Compare coefficients:

3A= 3⇒A=1,3B−A= 8⇒3B−1=8, B=3,3C−B= −21⇒3C−3=−21, C=−6,−C+R= −25⇒6+R=−25, R=−31.\begin{align*} 3A=&\,3 &&\Rightarrow A=1,\\ 3B-A=&\,8 &&\Rightarrow 3B-1=8,\ B=3,\\ 3C-B=&\,-21 &&\Rightarrow 3C-3=-21,\ C=-6,\\ -C+R=&\,-25 &&\Rightarrow 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*}