A-Level Maths: Applications of the Factor Theorem | A-Level数学：因式定理的应用

📚 A-Level Maths: Applications of the Factor Theorem | A-Level数学：因式定理的应用

The Factor Theorem is one of the most powerful tools in A-Level algebra. It allows us to find linear factors of polynomials quickly, and from there we can factorise cubic and higher-order expressions, solve equations, sketch graphs, and even determine unknown coefficients. This article explains the theorem, its key applications, and typical exam-style questions with full solutions.

因式定理是A-Level代数中最有力的工具之一。它能帮助我们快速找出多项式的线性因式，从而进一步对三次及更高次多项式进行因式分解、解方程、绘制图像，甚至求出未知系数。本文将讲解这一定理的核心内容、主要应用以及典型考试题型的完整解答。

1. What Is the Factor Theorem? | 什么是因式定理？

For a polynomial f(x), if f(a) = 0, then (x − a) is a factor of f(x). Conversely, if (x − a) is a factor of f(x), then f(a) = 0. This statement is known as the Factor Theorem.

对于多项式 f(x)，若 f(a) = 0，则 (x − a) 是 f(x) 的一个因式；反过来，若 (x − a) 是 f(x) 的一个因式，则 f(a) = 0。这一结论被称为因式定理。

f(a) = 0 ⇔ (x − a) is a factor of f(x)

f(a) = 0 ⇔ (x − a) 是 f(x) 的因式

The theorem is a special case of the Remainder Theorem: when f(x) is divided by (x − a), the remainder is f(a). If the remainder is zero, the division is exact and (x − a) is a factor.

因式定理是余数定理的特例：f(x) 除以 (x − a) 所得的余数为 f(a)。当余数为零时，除法能整除，(x − a) 即为因式。

2. Testing Possible Roots | 检验可能的根

To apply the Factor Theorem, we first need candidate values of a to test. For a polynomial with integer coefficients, any integer root must divide the constant term. For example, for f(x) = 2x³ − 3x² − 11x + 6, the constant term is 6, so possible integer roots are ±1, ±2, ±3, ±6.

要应用因式定理，我们首先需要确定候选的 a 值。对于整数系数的多项式，任何整数根都必须是常数项的因数。例如，对于 f(x) = 2x³ − 3x² − 11x + 6，常数项为 6，因此可能的整数根为 ±1、±2、±3、±6。

In addition, rational roots of the form p/q must have p dividing the constant term and q dividing the leading coefficient. This is sometimes called the Rational Root Theorem and is an excellent way to narrow down candidates.

此外，形如 p/q 的有理根必须满足 p 整除常数项、q 整除首项系数。这一结论有时被称为有理根定理，是缩小候选范围的有效方法。

List factors of the constant term to find integer candidates.
列出常数项的因数，找出整数候选值。
Include both positive and negative candidates.
候选值要同时包含正数和负数。
If the leading coefficient is not 1, also consider fractional candidates p/q.
若首项系数不为 1，还要考虑分数候选值 p/q。

3. Factorising a Cubic Polynomial Completely | 三次多项式的完整因式分解

To factorise a cubic f(x), we start by finding one root a such that f(a) = 0. Then (x − a) is a factor, and we divide f(x) by (x − a) to obtain a quadratic, which we factorise using standard methods.

要对三次多项式 f(x) 进行因式分解，我们首先找到一个根 a 使得 f(a) = 0。于是 (x − a) 是一个因式，将 f(x) 除以 (x − a) 得到一个二次式，再用常规方法分解该二次式即可。

Worked Example | 例题：Factorise f(x) = x³ − 6x² + 11x − 6.

Test the factors of −6: f(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor.

检验 −6 的因数：f(1) = 1 − 6 + 11 − 6 = 0，所以 (x − 1) 是一个因式。

Divide f(x) by (x − 1): the quotient is x² − 5x + 6. Then factorise the quadratic: x² − 5x + 6 = (x − 2)(x − 3).

用 (x − 1) 除 f(x)：商为 x² − 5x + 6。再分解二次式：x² − 5x + 6 = (x − 2)(x − 3)。

x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3)

Notice that we found three linear factors, all of which come from integer roots. Whenever possible, express the polynomial as a product of linear factors and/or irreducible quadratics.

注意我们找到了三个线性因式，它们都来自整数根。在可能的情况下，应把多项式表示成线性因式与（或）不可约二次式的乘积。

4. Solving Polynomial Equations | 解多项式方程

Once a polynomial is fully factorised, solving f(x) = 0 becomes immediate: set each linear factor equal to zero and solve.

一旦多项式被完全因式分解，解方程 f(x) = 0 就变得十分直接：令每个线性因式等于零并求解即可。

Worked Example | 例题：Solve 2x³ − 3x² − 11x + 6 = 0.

Test candidates: f(3) = 54 − 27 − 33 + 6 = 0, so (x − 3) is a factor. Dividing gives:

检验候选值：f(3) = 54 − 27 − 33 + 6 = 0，因此 (x − 3) 是因式。相除得：

2x³ − 3x² − 11x + 6 = (x − 3)(2x² + 3x − 2)

Then 2x² + 3x − 2 = (2x − 1)(x + 2). Hence the roots are x = 3, x = ½, and x = −2.

又 2x² + 3x − 2 = (2x − 1)(x + 2)。因此根为 x = 3、x = ½ 和 x = −2。

In an exam, always state the full solution set and check that the number of roots matches the degree of the polynomial (counting multiplicity).

考试中，务必写出完整的解集，并核验根的个数是否与多项式次数一致（计重根）。

5. Finding Unknown Coefficients | 求未知系数

A very common examination question gives you a polynomial containing an unknown constant, together with the information that (x − a) is a factor. Substituting x = a into f(x) and setting the result to zero produces a linear or quadratic equation in the unknown coefficient.

一类非常常见的考题给出一个含有未知常数的多项式，并告知 (x − a) 是其因式。将 x = a 代入 f(x) 并令结果为零，即可得到关于未知系数的一次或二次方程。

Worked Example | 例题：Given that (x − 1) is a factor of f(x) = x³ + kx² + 2x − 4, find the value of k.

解：已知 (x − 1) 是 f(x) = x³ + kx² + 2x − 4 的因式，求 k 的值。

Using the Factor Theorem: f(1) = 0. So 1 + k + 2 − 4 = 0, which gives k = 1.

由因式定理：f(1) = 0。因此 1 + k + 2 − 4 = 0，解得 k = 1。

The same idea extends to two unknowns. If two factors are given, substitute both roots to obtain two simultaneous equations and solve them.

同样的思路可推广到两个未知数：若给出两个因式，将两个根分别代入即可得到两个联立方程并求解。

Extended Example | 进阶例题：f(x) = x³ + px² + qx − 6 has factors (x − 1) and (x + 2). Find p and q.

f(1) = 1 + p + q − 6 = 0 ⇒ p + q = 5. f(−2) = −8 + 4p − 2q − 6 = 0 ⇒ 4p − 2q = 14 ⇒ 2p − q = 7. Solving gives p = 4 and q = 1.

f(1) = 1 + p + q − 6 = 0，即 p + q = 5；f(−2) = −8 + 4p − 2q − 6 = 0，即 4p − 2q = 14，亦即 2p − q = 7。联立解得 p = 4，q = 1。

6. The Factor Theorem and Graphs | 因式定理与函数图像

Each linear factor (x − a) of f(x) corresponds to an x-intercept at x = a on the graph of y = f(x). Therefore, finding factors helps us sketch polynomial graphs accurately.

f(x) 的每个线性因式 (x − a) 都对应 y = f(x) 图像上的一个 x 轴截距 x = a。因此，找出因式有助于准确绘制多项式图像。

For f(x) = x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3), the graph crosses the x-axis at 1, 2, and 3. Since the leading coefficient is positive, the graph falls from the left and rises to the right, with the characteristic S-shape of a cubic.

对于 f(x) = x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3)，图像在 x = 1、2、3 处穿过 x 轴。由于首项系数为正，图像从左下方延伸至右上方，呈现三次函数的典型 S 形。

Distinct real roots give distinct x-intercepts.
互不相同的实根对应不同的 x 轴截距。
A repeated root (x − a)² touches the axis but does not cross it.
重根 (x − a)² 处图像与 x 轴相切但不穿过。
A cubic with three real roots can be sketched using only the intercepts and the end behaviour.
具有三个实根的三次函数仅凭截距和端部行为即可作图。

7. Repeated Factors and Their Meaning | 重因式及其含义

If (x − a) is a repeated factor, the Factor Theorem alone tells us f(a) = 0, but the graph behaves differently: it touches the x-axis at a rather than crossing it. In Edexcel A-Level, you may also use the result that (x − a)² divides f(x) when f(a) = 0 and f′(a) = 0.

若 (x − a) 是重因式，仅靠因式定理只能得到 f(a) = 0，但图像行为不同：它在 a 处与 x 轴相切而非穿过。在爱德思A-Level中，还可以利用如下结论：当 f(a) = 0 且 f′(a) = 0 时，(x − a)² 整除 f(x)。

Worked Example | 例题：Show that (x − 2)² is a factor of f(x) = x³ − 5x² + 8x − 4.

解：证明 (x − 2)² 是 f(x) = x³ − 5x² + 8x − 4 的因式。

f(2) = 8 − 20 + 16 − 4 = 0. Also f′(x) = 3x² − 10x + 8, so f′(2) = 12 − 20 + 8 = 0. Hence (x − 2)² is a factor. In fact f(x) = (x − 2)²(x − 1).

f(2) = 8 − 20 + 16 − 4 = 0；又 f′(x) = 3x² − 10x + 8，所以 f′(2) = 12 − 20 + 8 = 0。因此 (x − 2)² 是因式。事实上 f(x) = (x − 2)²(x − 1)。

8. Linking with the Remainder Theorem | 与余数定理的联系

The Remainder Theorem states that when f(x) is divided by (x − a), the remainder is f(a). The Factor Theorem is simply the special case where this remainder equals zero. In exam questions, you may need to combine both ideas.

余数定理指出：f(x) 除以 (x − a) 的余数为 f(a)。因式定理不过是余数为零的特例。考试题目常常需要同时运用这两个定理。

Remainder Theorem \| 余数定理	Factor Theorem \| 因式定理
Remainder = f(a) when dividing by (x − a)	Remainder = 0, so (x − a) is a factor
除以 (x − a) 的余数为 f(a)	余数为 0，故 (x − a) 是因式

For example, if f(x) leaves a remainder of 5 when divided by (x − 2), then f(2) = 5. If it is divisible by (x + 1), then f(−1) = 0. Always determine whether the question asks for a remainder or for divisibility before substituting.

例如，若 f(x) 除以 (x − 2) 的余数为 5，则 f(2) = 5；若它能被 (x + 1) 整除，则 f(−1) = 0。代入前务必看清题目问的是余数还是整除关系。

9. Exam-Style Worked Example | 考试题型示例

Question | 题目：f(x) = 2x³ + ax² − 7x + b. Given that (x − 1) is a factor and that f(x) leaves a remainder of 20 when divided by (x + 2), find a and b, and hence factorise f(x) completely.

解答：f(x) = 2x³ + ax² − 7x + b。已知 (x − 1) 是因式，且 f(x) 除以 (x + 2) 的余数为 20，求 a 和 b，并完整因式分解 f(x)。

Since (x − 1) is a factor: f(1) = 2 + a − 7 + b = 0 ⇒ a + b = 5.

因为 (x − 1) 是因式：f(1) = 2 + a − 7 + b = 0，即 a + b = 5。

Since the remainder is 20 on division by (x + 2): f(−2) = −16 + 4a + 14 + b = 20 ⇒ 4a + b = 22.

除以 (x + 2) 余数为 20：f(−2) = −16 + 4a + 14 + b = 20，即 4a + b = 22。

Solving simultaneously: a = 17/3, b = −2/3. Then f(x) = 2x³ + (17/3)x² − 7x − 2/3. Dividing by (x − 1) gives f(x) = (x − 1)(2x² + (23/3)x + 2/3). Factorising the quadratic: (3x + 2)(2x + 1)/3, so the full factorisation is f(x) = (x − 1)(2x + 1)(3x + 2)/3.

联立求解：a = 17/3，b = −2/3。于是 f(x) = 2x³ + (17/3)x² − 7x − 2/3。除以 (x − 1) 得 f(x) = (x − 1)(2x² + (23/3)x + 2/3)。分解二次式得 (3x + 2)(2x + 1)/3，所以完整因式分解为 f(x) = (x − 1)(2x + 1)(3x + 2)/3。

10. Common Pitfalls and Exam Tips | 常见错误与应考技巧

Students often lose marks by testing only positive values, forgetting that negative roots are just as likely. Always test factors of the constant term including negatives. Another frequent mistake is substituting the wrong sign: for the factor (x + 3), you must test x = −3, not x = 3.

考生常因只检验正值而失分——负根同样常见。务必检验常数项的因数，包括负数。另一个常见错误是代入符号搞错：(x + 3) 是因式时应代 x = −3，而不是 x = 3。

Always write f(a) = 0 before stating the factor.
先写出 f(a) = 0，再给出因式结论。
Check every candidate systematically: it is acceptable to list a few failed tests in your working.
系统检验每个候选值：在草稿中列出几个失败测试是可以的。
When dividing, double-check your arithmetic; a small slip changes the quadratic completely.
做除法时反复检查运算；一个小失误会完全改变二次式。
If a quadratic factor does not factorise, leave it in the form ax² + bx + c and use the quadratic formula when solving.
若二次因式不能分解，保留 ax² + bx + c 的形式，求解时使用求根公式。

11. Practice Questions | 巩固练习

1. Show that (x + 2) is a factor of f(x) = x³ + 4x² + x − 6 and factorise f(x) completely.

1. 证明 (x + 2) 是 f(x) = x³ + 4x² + x − 6 的因式，并完整因式分解。

2. Given that (x − 3) is a factor of g(x) = 2x³ − 9x² + kx − 3, find k and solve g(x) = 0.

2. 已知 (x − 3) 是 g(x) = 2x³ − 9x² + kx − 3 的因式，求 k 并解方程 g(x) = 0。

3. The polynomial h(x) = x³ + px² + qx + 8 has a remainder of 20 when divided by (x − 1) and a remainder of −8 when divided by (x + 2). Find p and q.

3. 多项式 h(x) = x³ + px² + qx + 8 除以 (x − 1) 的余数为 20，除以 (x + 2) 的余数为 −8。求 p 和 q。

Answers: 1. (x + 2)(x − 1)(x + 3); 2. k = 3, roots x = 3, x = ½, x = 1; 3. p = 3, q = −2.

参考答案：1. (x + 2)(x − 1)(x + 3)；2. k = 3，根为 x = 3、x = ½、x = 1；3. p = 3，q = −2。

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A-Level Maths: Applications of the Factor Theorem | A-Level数学：因式定理的应用