📚 Example 1.7.2: Solving a Cubic by Factorisation | 例1.7.2：用因式分解法求解三次方程
In this worked example, we explore how to solve a cubic equation using the Factor Theorem and polynomial division. This technique is a core requirement in A-Level mathematics, especially when a root is already provided. We will go through each step carefully, from verification to the final solution set.
在本例中，我们将学习如何运用因式定理和多项式除法来求解三次方程。这是A-Level数学中的核心考点，尤其当已知一个根时。我们将细致地从验证根开始，逐步推导出完整解集。
1. Introduction to the Problem | 问题引入
the cubic equation 2x³ − 3x² − 8x + 12 = 0, and we are told that x = 2 is a root. The word “root” = 2 into the equation produces zero. Our goal is to find all three roots of the equation.
题目要求我们求解三次方程 2x³ − 3x² − 8x + 12 = 0，并已知 x = 2 是它的一个根。“根”的含义是：将 x = 2 代入方程后结果为 0。我们的目标是找出该方程的三个根。
2. Verifying the Given Root | 验证已知根
t-hand side of the equation to confirm that it indeed equals zero. This is a crucial first step because it validates the use of the Factor Theorem.
首先，我们将 x = 2 代入等号左边的表达式，确认其值确实为 0。这是关键的第一步，因为它是我们应用因式定理的前提。
2(2)³ − 3(2)² − 8(2) + 12 = 2×8 − 3×4 − 16 + 12 = 16 − 12 − 16 + 12 = 0
rmed as a root. Therefore, by the Factor Theorem, (x − 2) must be a factor of the cubic expression.
由于结果为零，x = 2 被确认为一个根。因此，根据因式定理，(x − 2) 必定是该三次表达式的一个因式。
3. Understanding the Factor Theorem | 理解因式定理
orem is a direct consequence of the Remainder Theorem and is extremely useful for factorising higher-degree polynomials.
因式定理指出：对于多项式 f(x)，如果 f(a) = 0，那么 (x − a) 是 f(x) 的因式；反过来，如果 (x − a) 是因式，则 f(a) = 0。该定理是余数定理的直接推论，在分解高次多项式时非常有用。
− 8x + 12 exactly, leaving a quadratic quotient. Our next task is to find that quadratic.
在我们的例子中，f(2) = 0 意味着 (x − 2) 可以整除 f(x) = 2x³ − 3x² − 8x + 12，余数为零，得到一个二次商式。下一步就是求出该二次式。
4. Polynomial Division | 多项式除法
We divide f(x) by (x − 2). Write the division as a long-division process, ensuring all terms are in descending powers of x. The dividend is 2x³ − 3x² − 8x + 12, and the divisor is x − 2.
我们用 (x − 2) 去除 f(x)。将长除法过程写出来，确保所有项按 x 的降幂排列。被除数是 2x³ − 3x² − 8x + 12，除数是 x − 2。
by 2x² to get 2x³ − 4x². Subtract from the dividend to get (2x³ − 3x²) − (2x³ − 4x²) = x².
第一步：用首项 2x³ 除以 x 得 2x²。将 (x − 2) 乘以 2x² 得到 2x³ − 4x²。从被除数中减去，得到 (2x³ − 3x²) − (2x³ − 4x²) = x²。
get x. Multiply (x − 2) by x to get x² − 2x. Subtract to get (x² − 8x) − (x² − 2x) = −6x.
第二步：将下一项 −8x 拉下来。用 x² 除以 x 得 x。将 (x − 2) 乘以 x 得到 x² − 2x。相减得到 (x² − 8x) − (x² − 2x) = −6x。
t −6. Multiply (x − 2) by −6 to get −6x + 12. Subtract to get (−6x + 12) − (−6x + 12) = 0.
第三步：将常数项 12 拉下来。用 −6x 除以 x 得 −6。将 (x − 2) 乘以 −6 得到 −6x + 12。相减得到 (−6x + 12) − (−6x + 12) = 0。
The quotient is therefore 2x² + x − 6. We can write the original expression as a product of the divisor and the quotient:
因此商式为 2x² + x − 6。我们可以将原表达式写为除数与商式的乘积：
2x³ − 3x² − 8x + 12 = (x − 2)(2x² + x − 6)
5. Quotient Factorisation | 商式因式分解
(the coefficient of x). The numbers are 4 and −3, because 4 × (−3) = −12 and 4 + (−3) = 1.
现在我们对二次商式 2x² + x − 6 进行因式分解。我们需要找到两个数，它们的乘积等于 2 × (−6) = −12，且它们的和等于 x 的系数 +1。这两个数是 4 和 −3，因为 4 × (−3) = −12 且 4 + (−3) = 1。
We split the middle term and factor by grouping:
我们使用拆项分组法进行因式分解：
2x² + x − 6 = 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2)
Thus the fully factorised form of the cubic is:
因此三次式的完全因式分解形式为：
2x³ − 3x² − 8x + 12 = (x − 2)(2x − 3)(x + 2)
6. Solving the Quadratic Factor | 解二次因式
Once the cubic is factorised into linear factors, we can set each factor equal to zero to find the roots. The original equation becomes:
当三次式被分解为线性因式的乘积后，我们令每个因式等于零，从而求出根。原方程变为：
(x − 2)(2x − 3)(x + 2) = 0
Using the zero-product property, if the product is zero, then at least one of the factors must be zero:
利用零乘积性质，若乘积为零，则至少有一个因式为零：
x − 2 = 0 → x = 2 (the given root)
x − 2 = 0 → x = 2（已知根）
2x − 3 = 0 → x = 3⁄2
2x − 3 = 0 → x = 3⁄2
x + 2 = 0 → x = −2
x + 2 = 0 → x = −2
7. Complete Solution Set | 完整解集
he solution set is {2, 3⁄2, −2}. We can also write the roots in ascending order: −2, 3⁄2, 2.
方程的三个根为 x = 2、x = 3⁄2 和 x = −2。用集合记号表示，解集为 {2, 3⁄2, −2}。按升序排列为：−2, 3⁄2, 2。
These roots can be checked individually by substituting back into the original equation, which we will do shortly. Notice that the degree of the polynomial is three, and we have found exactly three roots, counting multiplicities if any.
我们可以通过逐个代回原方程来检验这些根，稍后我们将进行操作。注意该多项式的次数为三，而我们已经找到恰好三个根（若计重数）。
8. Alternative Method: Synthetic Division | 替代方法：合成除法
r factor of the form (x − a). For our problem, since the divisor is (x − 2), we use a = 2.
当用形如 (x − a) 的线性因式去除多项式时，合成除法是比长除法更快捷的方法。对于本题，除数为 (x − 2)，因此取 a = 2。
o get 0. The last number is the remainder, which is zero, confirming that x = 2 is a root.
按顺序写出 f(x) 的系数：2, −3, −8, 12。将首个系数 2 拉下。乘以 a = 2 得 4，加到 −3 上得 1。将 1 乘以 2 得 2，加到 −8 上得 −6。将 −6 乘以 2 得 −12，加到 12 上得 0。最后一个数是余数，为 0，再次确认 x = 2 是根。
| 2 | −3 | −8 | 12 | |
| 2 | 4 | 2 | −12 | |
| 2 | 1 | −6 | 0 |
The numbers in the last row, except the final zero, give the coefficients of the quotient: 2x² + x − 6, exactly as we obtained earlier.
最后一行中除末尾零以外的数给出了商式的系数：2x² + x − 6，与我们之前得到的结果完全一致。
9. Checking the Answers | 检查答案
substitution. Let us check each root in the original equation f(x) = 2x³ − 3x² − 8x + 12.
验证每个根是良好的习惯。现在我们将每个根代入原方程 f(x) = 2x³ − 3x² − 8x + 12 进行检验。
For x = 2: 2(2)³ − 3(2)² − 8(2) + 12 = 16 − 12 − 16 + 12 = 0 ✓
当 x = 2 时：2(2)³ − 3(2)² − 8(2) + 12 = 16 − 12 − 16 + 12 = 0 ✓
or x = 3⁄2: 2(3⁄2)³ − 3(3⁄2)² − 8(3⁄2) + 12 = 2(27⁄8) − 3(9⁄4) − 12 + 12 = 27⁄4 − 27⁄4 = 0 ✓
当 x = 3⁄2 时：2(3⁄2)³ − 3(3⁄2)² − 8(3⁄2) + 12 = 2(27⁄8) − 3(9⁄4) − 12 + 12 = 27⁄4 − 27⁄4 = 0 ✓
For x = −2: 2(−2)³ − 3(−2)² − 8(−2) + 12 = 2(−8) − 3(4) + 16 + 12 = −16 − 12 + 16 + 12 = 0 ✓
当 x = −2 时：2(−2)³ − 3(−2)² − 8(−2) + 12 = 2(−8) − 3(4) + 16 + 12 = −16 − 12 + 16 + 12 = 0 ✓
All three values satisfy the equation, confirming that our factorisation and solutions are correct.
三个值都满足原方程，确认了我们的因式分解和解集都是正确的。
10. Common Mistakes to Avoid | 常见错误避免
When solving cubic equations with a given root, students often make several common errors. Here we highlight the most frequent ones and how to avoid them.
在利用已知根求解三次方程时，同学们常犯几个典型错误。下面我们指出最常见的几种及其规避方法。
given root before applying the Factor Theorem. Always substitute to confirm that f(a) = 0.
在应用因式定理之前忘记验证已知根。务必通过代入确认 f(a) = 0。
Making sign errors during polynomial long division, especially when subtracting negative terms. Write each line carefully and use brackets.
在多项式长除法中出现符号错误，尤其是减去负项时。注意每一步对齐，并使用括号。
Factorising the quadratic incorrectly. If the coefficient of x² is not 1, use the product-and-sum method or the AC method carefully.
对二次式进行因式分解出错。当 x² 的系数不为 1 时，应谨慎使用乘积与和法或 AC 法。
Stopping after finding only one root. A cubic must have three roots (counting multiplicity). Always factor completely.
找到一个根后就停止。三次方程必须有三个根（计重数）。务必彻底因式分解。
Mixing up the signs when using synthetic division. Remember to bring down, multiply, and add in the correct order.
使用合成除法时混淆符号。记住顺序：下拉、相乘、相加。
11. Connection to the AQA Specification | 与AQA考纲的联系
In AQA A-Level Mathematics, the Factor Theorem and polynomial division appear in the Pure Mathematics content. The ability to solve cubic equations by factorisation is a fundamental skill, and questions often provide one root to start the process. This example 1.7.2 illustrates the exact style of problem you may encounter in an AS or A-level exam.
在AQA A-Level数学的纯数学部分中，因式定理和多项式除法是重要考点。通过因式分解求解三次方程是一项基本技能，题目通常会给出一个根作为切入点。本例题 1.7.2 展示了你在AS或A-level考试中可能遇到的典型题型。
When working with such problems, always write down the Factor Theorem clearly, show all polynomial division steps, and present your final factorised form. Examiners award marks for method and accuracy, so a well-structured solution is essential.
处理此类题目时，务必清晰地写出因式定理，展示多项式除法的所有步骤，并写出最终的因式分解形式。考官会根据方法和准确性给分，因此结构清晰的解答至关重要。
12. Summary | 总结
)(x + 2), and then solved each linear factor to find the roots x = 2, x = 3⁄2, and x = −2.
我们已经成功运用以下步骤求解了三倍方程 2x³ − 3x² − 8x + 12 = 0：验证已知根 x = 2，将多项式除以 (x − 2) 得到二次式，将二次式分解为 (2x − 3)(x + 2)，然后令每个线性因式为零，求得根 x = 2、x = 3⁄2 和 x = −2。
This example demonstrates the power of the Factor Theorem and highlights the importance of tidy algebra. By mastering this technique, you will be well-prepared for similar questions in your AQA examinations.
本例展示了因式定理的强大威力，也凸显了整洁代数运算的重要性。掌握这一方法后，你将能够从容应对AQA考试中的同类问题。
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