Solve cubic equations using factorisation or other appropriate methods

Objective

• Apply the Remainder and Factor Theorems to cubic polynomials.
• Generate and test all possible rational roots using the Rational Root Theorem.
• Factor a cubic by grouping, synthetic (or long) division, or by recognising a depressed form.
• Solve the remaining quadratic factor and verify every solution.

1. Syllabus Alignment (Cambridge Additional Mathematics 0606)

Syllabus Outcome	What the Notes Cover	Suggested Improvement
3.1 Remainder & Factor Theorems	Full statement of the Remainder Theorem with a worked example.	Add a “quick‑check” checklist that students can copy onto a formula sheet.
3.2 Find factors of polynomials	Rational‑Root Theorem, systematic candidate list, reduction of fractions.	Provide a compact table of common candidate sets for leading coefficients 1–6 and stress testing both signs.
3.3 Solve cubic equations	Three methods (grouping, Rational‑Root + synthetic division, optional depressed cubic) with a decision‑tree flow and several worked examples.	Place the decision‑flow diagram directly after the methods list; add a reminder that “if no rational root is found, the cubic has no rational solution”.

2. Quick‑Check Formula Sheet (Remainder & Factor Theorems)

1. Write the polynomial in the form P(x).
2. Plug the candidate k into P(k).
3. If P(k)=0 → (x‑k) is a factor (Factor Theorem).
   Otherwise the remainder is P(k) and (x‑k) is not a factor.

Copy this three‑step checklist onto the back of your formula sheet for rapid use.

3. Consistent Notation

Throughout the notes k denotes the numeric root. The corresponding linear factor is written as (x‑k). If the root is negative, e.g. k = –2, the factor becomes (x + 2). Remember: the sign inside the bracket is the opposite of the root value.

4. Review of Key Concepts

4.1 Polynomials and Degree

A polynomial in x is

\[ P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0,\qquad a_neq0, \] where n is the highest power (the degree).

4.2 Remainder Theorem (with checklist)

If P(x) is divided by (x‑k) the remainder is P(k). Consequently:

• P(k)=0 → remainder 0 → (x‑k) is a factor (Factor Theorem).
• P(k)≠0 → remainder P(k) → (x‑k) is not a factor.

4.3 Rational Root Theorem

For a polynomial with integer coefficients, any rational root \(\displaystyle \frac{p}{q}\) (in lowest terms) satisfies

• \(p\) divides the constant term \(a_0\).
• \(q\) divides the leading coefficient \(a_n\).

To generate the complete list of candidates:

1. List all positive factors of \(|a_0|\) → p‑list.
2. List all positive factors of \(|a_n|\) → q‑list.
3. Form every fraction \(\pm\frac{p}{q}\) and reduce to lowest terms.
4. Remember to test **both** the positive and negative versions of every reduced fraction.

4.4 Compact Candidate‑Set Table (Typical Leading Coefficients)

Leading Coefficient \(a_n\)	Typical Candidate Set \(\pm\frac{p}{q}\)
1	\(\pm\) factors of \(a_0\)
2	\(\pm\) factors of \(a_0\) and \(\pm\frac{\text{factors of }a_0}{2}\)
3	\(\pm\) factors of \(a_0\), \(\pm\frac{\text{factors of }a_0}{3}\)
4	\(\pm\) factors of \(a_0\), \(\pm\frac{\text{factors of }a_0}{2}\), \(\pm\frac{\text{factors of }a_0}{4}\)
5	\(\pm\) factors of \(a_0\), \(\pm\frac{\text{factors of }a_0}{5}\)
6	\(\pm\) factors of \(a_0\), \(\pm\frac{\text{factors of }a_0}{2}\), \(\pm\frac{\text{factors of }a_0}{3}\), \(\pm\frac{\text{factors of }a_0}{6}\)

Copy this table onto your revision sheet for rapid reference.

5. Methods for Solving Cubic Equations

Method	When to Use	Key Steps	Typical Example
5.1 Factor by Grouping	The terms can be split into groups that share a common factor.	Rewrite the cubic as two (or three) groups. Factor each group. Factor out the common binomial. Solve the resulting linear and quadratic factors.	\(3x^3+6x^2-2x-4=0\)
5.2 Rational‑Root Test + Factor Theorem + Synthetic Division	General cubic where grouping fails.	List every possible rational root \(\pm\frac{p}{q}\) (use the candidate‑set table). Apply the quick‑check (Remainder Theorem) to each candidate. Stop when P(k)=0. Factor out \((x‑k)\) using synthetic (or long) division to obtain a quadratic. Solve the quadratic by factoring, completing the square, or the quadratic formula.	\(2x^3-3x^2-8x+12=0\)
5.3 Depressed Cubic (Cardano) – optional	Advanced A‑Level work; not required for IGCSE.	Substitute \(x = y-\dfrac{b}{3a}\) to remove the quadratic term. Obtain a depressed cubic \(y^3+py+q=0\). Apply Cardano’s formula.	Beyond the scope of this syllabus.

5.4 Decision‑Flow Diagram (see visual aid at the end)

When you encounter a cubic, follow the flowchart to decide which method is most efficient. The diagram is placed after the practice section for easy reference.

6. Detailed Worked Examples

6.1 Example – Factor by Grouping

\[ 3x^3+6x^2-2x-4=0 \] \[ \begin{aligned} (3x^3+6x^2)&+(-2x-4)=0\\ 3x^2(x+2)&-2(x+2)=0\\ (x+2)(3x^2-2)&=0 \end{aligned} \] \[ \Rightarrow x=-2\quad\text{or}\quad 3x^2-2=0\;\Longrightarrow\;x=\pm\sqrt{\frac{2}{3}}. \]

6.2 Example – Rational‑Root Test, Factor Theorem & Synthetic Division

Solve 2x³‑3x²‑8x + 12 = 0.

1. Possible rational roots (using the candidate‑set table for \(a_n=2\)):
   • Factors of constant term 12: \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\).
   • Factors of leading coefficient 2: \(\pm1,\pm2\).
   • Candidates \(\displaystyle \pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac12,\pm\frac32\) (duplicates removed).
2. Quick‑check (Factor Theorem):
   • P(2)=2·8‑3·4‑8·2+12=0 → root found: \(k=2\).
3. Synthetic division by \((x‑2)\):

      2 | 2   -3   -8   12
          |      4    2   -12
          --------------------
            2    1   -6    0
           

   Quadratic factor: \(2x^2 + x - 6\).
4. Solve the quadratic: \[ x=\frac{-1\pm\sqrt{1+48}}{4}=\frac{-1\pm7}{4}\Longrightarrow x=\frac{6}{4}=1.5\;\text{or}\;x=-2. \]
5. All solutions: \(\boxed{x=2,\;1.5,\;-2}\).

Common‑Pitfall Box (Synthetic Division)

6.3 Example – Cubic Inequality (link to syllabus 4.5)

Solve \(\displaystyle 2x^3-3x^2-8x+12\ge 0\).

1. Find the zeros using the method of 6.2 (roots: \(-2,\;1.5,\;2\)).
2. Arrange them in increasing order: \(-2 < 1.5 < 2\).
3. Test the sign of the cubic in each interval:
   • For \(x<-2\): pick \(x=-3\) → \(P(-3)= -54-27+24+12=-45 <0\).
   • For \(-20\).
   • For \(1.5
   • For \(x>2\): pick \(x=3\) → \(P(3)=54-27-24+12=15>0\).
4. Include the zeros because the inequality is “\(\ge\)”.
5. Solution set: \[ x\in[-2,\;1.5]\;\cup\;[2,\;\infty). \]

7. Additional Worked Examples (Covering All Syllabus Points)

1. Remainder Theorem in action Find the remainder when \(P(x)=x^3-4x^2+5x-2\) is divided by \((x-2)\). \(P(2)=8-16+10-2=0\) → remainder 0, so \((x-2)\) is a factor.
2. Factor by grouping (different numbers) \(4x^3+4x^2-15x-15=0\) \((4x^2-15)(x+1)=0\) → \(x=-1,\;x=\pm\sqrt{15/4}\).
3. Integer leading coefficient \(x^3-6x^2+11x-6=0\). Candidates \(\pm1,\pm2,\pm3,\pm6\); \(P(1)=0\). Synthetic division → \(x^2-5x+6\) → \((x-2)(x-3)\). Solutions: \(1,2,3\).
4. Non‑unit leading coefficient \(6x^3+5x^2-12x-10=0\). Candidates from the table for \(a_n=6\) give \(\pm1,\pm2,\pm5,\pm\frac12,\pm\frac53,\pm\frac{5}{3},\pm\frac{5}{6},\pm\frac{10}{3}\). Testing shows \(x=1\) is a root. Synthetic division → \(6x^2- x -10\). Quadratic formula: \(x=\dfrac{1\pm\sqrt{241}}{12}\). Solutions: \(1,\; \dfrac{1\pm\sqrt{241}}{12}\).
5. Factor‑theorem verification before division Verify \((x+2)\) for \(4x^3+8x^2-5x-10\). \(P(-2)=0\) → factor confirmed. Synthetic division → \(4x^2-5\). Remaining roots: \(x=-2,\;x=\pm\frac{\sqrt5}{2}\).

8. Common Pitfalls & How to Avoid Them

• Skipping negative candidates: Always write both \(\pm\) for each reduced fraction.
• Not reducing fractions: Reduce \(\frac{6}{4}\) to \(\frac{3}{2}\) before testing.
• Sign errors in synthetic division: Remember the divisor is \(-k\) when the factor is \((x+k)\).
• Assuming an integer root exists: If none of the integer candidates work, move on to fractional candidates.
• Forgetting to solve the quadratic factor: After extracting a linear factor, always finish the problem.
• Not checking the final answers: Substitute each root back into the original cubic.

9. Practice Questions

1. Solve \(x^3-4x^2-7x+10=0\) using the Rational Root Theorem.
2. Factor completely: \(6x^3+5x^2-12x-10\).
3. Find all real solutions of \(2x^3-9x^2+12x-4=0\) (show the full Rational Root Test).
4. Given that \(x=3\) is a root of \(x^3+ax^2+bx-27=0\), determine \(a\) and \(b\) and solve the cubic completely.
5. Show that \(x=-2\) is a root of \(4x^3+8x^2-5x-10=0\) and solve the remaining quadratic factor.
6. Using the Remainder Theorem, find the remainder when \(P(x)=2x^3-5x^2+7x-3\) is divided by \((x+1)\). Is \((x+1)\) a factor?
7. For the cubic \(3x^3-2x^2-8x+4=0\), list all possible rational roots, test them, and obtain the complete set of solutions.
8. Solve the inequality \(2x^3-3x^2-8x+12\ge0\) (use the roots found in example 6.2).

10. Summary Checklist (What to Do When You See a Cubic)

1. Write the equation in standard form \(ax^3+bx^2+cx+d=0\).
2. If a factor is already suggested, apply the quick‑check checklist (Remainder Theorem).
3. Generate the full list of rational candidates using the candidate‑set table.
4. Test each candidate (both signs) with the quick‑check; stop at the first root.
5. Factor out \((x‑k)\) by synthetic (or long) division.
6. Solve the resulting quadratic (factor, complete the square, or quadratic formula).
7. Check every solution by substitution.
8. If no rational root is found, conclude that the cubic has no rational solution (Cardano not required for IGCSE).

11. Decision‑Flow Diagram (Visual Aid)

12. Final Note

Mastering cubic equations relies on a systematic approach: generate candidates, use the quick‑check, divide correctly, and always finish by solving the quadratic. Keep the quick‑check checklist, the candidate‑set table, and the decision‑flow diagram handy—they are the three tools that will make the process almost automatic.