题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

$f(x) = 4x^3 + ax^2 - 29x + b$

where $a$ and $b$ are constants.

Given that $(2x + 1)$ is a factor of $f(x)$ ,

(a) show that $a + 4b = -56$ .

Given also that when $f(x)$ is divided by $(x - 2)$ the remainder is $-25$ ,

(b) find a second simplified equation linking $a$ and $b$ .

(i) find the value of $a$ and the value of $b$ ,

(ii) fully factorise $f(x)$ .

(9)

题目中文翻译

本题必须展示所有解题步骤，不能完全依赖计算器技术。

$f(x) = 4x^3 + ax^2 - 29x + b$ ，其中 $a$ 、 $b$ 为常数。

已知 $(2x + 1)$ 是 $f(x)$ 的因式，

(a) 证明 $a + 4b = -56$ 。

又已知 $f(x)$ 除以 $(x - 2)$ 的余数为 $-25$ ，

(b) 求另一个关于 $a$ 和 $b$ 的简化方程。

(i) 求 $a$ 和 $b$ 的值，

(ii) 完全因式分解 $f(x)$ 。

（9分）

解答

利用因式定理：若 $(2x + 1)$ 是 $f(x)$ 的因式，则 $f\left(-\dfrac{1}{2}\right) = 0$ 。结合余式定理 $f(2) = -25$ ，联立方程组求解。

(a) By the Factor Theorem, since $(2x + 1)$ is a factor of $f(x)$ , we have $f\left(-\dfrac{1}{2}\right) = 0$ :

f\left(-\frac{1}{2}\right) =&\, 4\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right)^2 - 29\left(-\frac{1}{2}\right) + b \\[2mm] =&\, -\frac{1}{2} + \frac{a}{4} + \frac{29}{2} + b \\[2mm] =&\, \frac{a}{4} + 14 + b = 0 \end{align*}$$ $$a + 4b = -56 \quad \cdots (1)$$ **(b)** By the Remainder Theorem, since the remainder is $-25$ when $f(x)$ is divided by $(x - 2)$: $$\begin{align*} f(2) =&\, 4(2)^3 + a(2)^2 - 29(2) + b \\ =&\, 32 + 4a - 58 + b = -25 \end{align*}$$ $$4a + b = 1 \quad \cdots (2)$$ **(c)(i)** Solving equations $(1)$ and $(2)$ simultaneously: From $(2)$: $b = 1 - 4a$ Substituting into $(1)$: $$\begin{align*} a + 4(1 - 4a) =&\, -56 \\ a + 4 - 16a =&\, -56 \\ -15a =&\, -60 \\ a =&\, 4 \end{align*}$$ $$b = 1 - 4(4) = -15$$ $$\boxed{a = 4, \quad b = -15}$$ **(c)(ii)** $f(x) = 4x^3 + 4x^2 - 29x - 15$ Since $(2x + 1)$ is a factor, perform polynomial division: $$\begin{align*} 4x^3 + 4x^2 - 29x - 15 =&\, (2x + 1)(2x^2 + x - 15) \\ =&\, (2x + 1)(2x - 5)(x + 3) \end{align*}$$ $$\boxed{f(x) = (2x + 1)(2x - 5)(x + 3)}$$