📚 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？
(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
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
hen (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。
: 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
rised, 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
ing 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
raph 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。
(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
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。
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 |
termine 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： ) 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。
dratic: (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
e 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。
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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