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IAL 2025 Jan Q5

A Level / Edexcel / P2

IAL 2025 Jan Paper · Question 5

题目

Problem

f(x)=3x3+ax2−10x+bf(x)=3x^3+ax^2-10x+b

where aa and bb are constants.

Given that (3x−4)(3x-4) is a factor of f(x)f(x),

(a) show that 16a+9b=5616a+9b=56

(2)

Given further that when f(x)f(x) is divided by (x−2)(x-2) the remainder is bb,

(b) find the value of aa and the value of bb.

(4)

(c) Hence, using algebra, fully factorise f(x)f(x).

(3)

解答

(a)

解法一

思路

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(3x−4)(3x-4) 是因式，说明 x=43x=\frac43 是根，所以 f(43)=0f(\frac43)=0。

答题过程

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Since (3x−4)(3x-4) is a factor,

x=43\begin{align*} x=\frac43 \end{align*}

is a root. Therefore,

f(43)= 03(43)3+a(43)2−10(43)+b= 0.\begin{align*} f\left(\frac43\right)=&\,0\\ 3\left(\frac43\right)^3 +a\left(\frac43\right)^2 -10\left(\frac43\right)+b=&\,0. \end{align*}

So

649+16a9−403+b= 0.\begin{align*} \frac{64}{9}+\frac{16a}{9} -\frac{40}{3}+b=&\,0. \end{align*}

Multiply by 99:

64+16a−120+9b= 016a+9b= 56.\begin{align*} 64+16a-120+9b=&\,0\\ 16a+9b=&\,56. \end{align*}

This is the required result.

(b)

解法一

思路

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除以 (x−2)(x-2) 的余数是 f(2)f(2)。题目说余数是 bb，所以 f(2)=bf(2)=b。

答题过程

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By the remainder theorem,

f(2)=b.\begin{align*} f(2)=b. \end{align*}

Hence

3(2)3+a(2)2−10(2)+b= b24+4a−20+b= b4a+4= 0a= −1.\begin{align*} 3(2)^3+a(2)^2-10(2)+b=&\,b\\ 24+4a-20+b=&\,b\\ 4a+4=&\,0\\ a=&\,-1. \end{align*}

Using 16a+9b=5616a+9b=56,

16(−1)+9b= 569b= 72b= 8.\begin{align*} 16(-1)+9b=&\,56\\ 9b=&\,72\\ b=&\,8. \end{align*}

Therefore,

a=−1,b=8.\begin{align*} a=-1,\qquad b=8. \end{align*}

(c)

解法一

思路

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先代入 a=−1,b=8a=-1,b=8。由于题目要求 hence，要使用已知因式 (3x−4)(3x-4)，再把剩下的二次式分解。

答题过程

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Using a=−1a=-1 and b=8b=8,

f(x)=3x3−x2−10x+8.\begin{align*} f(x)=3x^3-x^2-10x+8. \end{align*}

Since (3x−4)(3x-4) is a factor,

3x3−x2−10x+8=  (3x−4)(x2+x−2).\begin{align*} 3x^3-x^2-10x+8 =&\,\, (3x-4)(x^2+x-2). \end{align*}

Now factorise the quadratic:

x2+x−2=  (x+2)(x−1).\begin{align*} x^2+x-2 =&\,\, (x+2)(x-1). \end{align*}

Therefore,

f(x)=(3x−4)(x+2)(x−1).\begin{align*} f(x)=(3x-4)(x+2)(x-1). \end{align*}

解法二

思路

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待定系数法（inspection）。已知 f(x)=3x3−x2−10x+8f(x)=3x^3-x^2-10x+8 并且 (3x−4)(3x-4) 是因式。设 f(x)=(3x−4)(ax2+bx+c)f(x) = (3x-4)(ax^2+bx+c)。通过两端的最高次项与常数项对比，一眼即可确定 a=1a=1 和 c=−2c=-2，然后再通过一次项或二次项求出 b=1b=1，能有效缩短计算时间。

答题过程

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Substitute a=−1a=-1 and b=8b=8:

f(x)=3x3−x2−10x+8\begin{align*} f(x) = 3x^3-x^2-10x+8 \end{align*}

Since (3x−4)(3x-4) is a factor of f(x)f(x), we can assume the factorised form is:

3x3−x2−10x+8=(3x−4)(ax2+bx+c)\begin{align*} 3x^3-x^2-10x+8 = (3x-4)(ax^2+bx+c) \end{align*}

Comparing coefficients:

  • For the x3x^3 term:
3a=3  ⟹  a=1\begin{align*} 3a = 3 \implies a = 1 \end{align*}
  • For the constant term:
−4c=8  ⟹  c=−2\begin{align*} -4c = 8 \implies c = -2 \end{align*}
  • For the x2x^2 term (expanding (3x−4)(x2+bx−2)(3x-4)(x^2+bx-2)):
3b−4a=−13b−4=−1……(since a=1)3b=  3b=  1\begin{align*} 3b - 4a =&-1\\[4mm] 3b - 4 =&-1\dots\dots(\text{since } a=1)\\[4mm] 3b =&\,\, 3\\[4mm] b =&\,\, 1 \end{align*}

Thus, the quadratic factor is (x2+x−2)(x^2+x-2).

Now, factorise the quadratic:

x2+x−2=(x+2)(x−1)\begin{align*} x^2+x-2 = (x+2)(x-1) \end{align*}

Therefore, the fully factorised form is:

f(x)=(3x−4)(x+2)(x−1)\begin{align*} f(x) = (3x-4)(x+2)(x-1) \end{align*}