How to Factorise a Polynomial of Degree 3
Factorising a polynomial of degree 3, often called a cubic polynomial, is a fundamental skill in algebra that opens the door to solving equations, graphing functions, and understanding higher‑level mathematics. Whether you are a student grappling with homework or a professional who needs to simplify expressions for engineering calculations, mastering the techniques to break down a cubic into linear and quadratic factors will greatly enhance your problem‑solving toolkit. This article walks you through the most reliable methods, provides a detailed step‑by‑step example, and answers common questions to ensure you can confidently factorise any cubic polynomial you encounter.
Introduction
When you are asked to factorise a polynomial of degree 3, you are essentially looking for three linear factors (or a combination of linear and quadratic factors) whose product equals the original expression. And the process typically involves identifying at least one rational root, using that root to reduce the cubic to a quadratic, and then applying familiar factoring techniques. On the flip side, by the end of this guide, you will understand how to apply the Rational Root Theorem, perform synthetic division, and use grouping or the quadratic formula as needed. The goal is not only to find the factors but also to verify that your work is correct, ensuring the final factorisation matches the original polynomial.
Key Concepts and Theorems
Before diving into the step‑by‑step procedures, it is helpful to review the underlying principles:
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Rational Root Theorem: Any possible rational root of a cubic polynomial (ax^3 + bx^2 + cx + d) is of the form (\frac{p}{q}), where (p) is a factor of the constant term (d) and (q) is a factor of the leading coefficient (a). This theorem narrows down the candidates you need to test It's one of those things that adds up..
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Factor Theorem: If (x = r) is a root of the polynomial, then ((x - r)) is a factor. Conversely, if ((x - r)) is a factor, then (x = r) is a root.
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Synthetic Division: A streamlined method for dividing a polynomial by a linear factor ((x - r)). It reduces the cubic to a quadratic quickly and with minimal arithmetic errors Still holds up..
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Quadratic Formula: Once you have reduced the cubic to a quadratic, you can factor the quadratic either by simple factoring or by applying the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}).
Step‑by‑Step Factoring Process
1. Write the Polynomial in Standard Form
Ensure the polynomial is written as (ax^3 + bx^2 + cx + d). As an example, consider the cubic:
[ 2x^3 - 7x^2 + 5x + 6 ]
2. List Possible Rational Roots
Apply the Rational Root Theorem:
- Factors of the constant term (d = 6): (\pm1, \pm2, \pm3, \pm6)
- Factors of the leading coefficient (a = 2): (\pm1, \pm2)
Possible rational roots (\frac{p}{q}) are:
[ \pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{3}{2} ]
3. Test the Candidates
Plug each candidate into the polynomial to see which yields zero. A quick way is to use synthetic division for each candidate. As an example, test (x = 2):
[ \begin{array}{r|rrrr} 2 & 2 & -7 & 5 & 6 \ & & 4 & -6 & -2 \ \hline & 2 & -3 & -1 & 4 \ \end{array} ]
The remainder is 4, so (x = 2) is not a root That's the part that actually makes a difference. Took long enough..
Test (x = -1):
[ \begin{array}{r|rrrr} -1 & 2 & -7 & 5 & 6 \ & & -2 & 9 & -14 \ \hline & 2 & -9 & 14 & -8 \ \end{array} ]
Remainder (-8); not a root Took long enough..
Continue until you find a zero remainder. In this example, (x = \frac{3}{2}) works:
[ \begin{array}{r|rrrr} \frac{3}{2} & 2 & -7 & 5 & 6 \ & & 3 & -3 & 3 \ \hline & 2 & -4 & 2 & 9 \ \end{array} ]
Oops, remainder 9. Let’s try (x = -2):
[ \begin{array}{r|rrrr} -2 & 2 & -7 & 5 & 6 \ & & -4 & 22 & -54 \ \hline & 2 & -11 & 27 & -48 \ \end{array} ]
Not zero. Finally, (x = 1):
[ \begin{array}{r|rrrr} 1 & 2 & -7 & 5 & 6 \ & & 2 & -5 & 0 \ \hline & 2 & -5 & 0 & 6 \ \end{array} ]
Remainder 6. The correct root is actually (x = -\frac{3}{2}):
[ \begin{array}{r|rrrr} -\frac{3}{2} & 2 & -7 & 5 & 6 \ & & -3 & 15 & -45 \ \hline & 2 & -10 & 20 & -39 \ \end{array} ]
Not zero. Let’s re‑evaluate: The polynomial (2x^3 - 7x^2 + 5x + 6) actually has integer roots. Testing (x = 3):
[ \begin{array}{r|rrrr} 3 & 2 & -7 & 5 & 6 \ & & 6 & -3 & 6 \ \hline & 2 & -1 & 2 & 12 \ \end{array} ]
Remainder 12. The correct root is (x = -\frac{3}{2}) after all? Let's double‑check with direct substitution:
(2(-\frac{3}{2})^3 - 7(-\frac{3}{2})^2 + 5(-\frac{3}{2}) + 6)
Calculate: ((- \frac{27}{4}) - \frac{63}{4} - \frac{15}{2} + 6 = -\frac{90}{4} - \frac{30}{4} + \frac{24}{4} = -\frac{96}{4} = -24). Not zero.
It appears the example polynomial may have been chosen arbitrarily. To keep the illustration clear, let’s switch to a polynomial that is guaranteed to have a simple rational root: (x^3 - 6x^2 + 11x - 6) Surprisingly effective..
4. Apply the Rational Root Theorem to the New Polynomial
- Constant term (d = -6): (\pm1, \pm2, \pm3, \pm6)
- Leading coefficient (a = 1): (\pm1, \pm2, \pm3, \pm6)
Thus, possible roots are (\pm1, \pm2, \pm3, \pm6).
