Quantitative Aptitude · Equations

Cubic and Higher-Degree Equations for CA Foundation

Updated 1 October 2026 · Fact-checked

A cubic equation has degree 3 and at most three roots. To solve it, test small factors of the constant term until f(a) = 0. By the factor theorem, (x − a) is then a factor. Divide it out by synthetic division, then solve the remaining quadratic for the other two roots.

Understand Cubic and Higher-Degree Equations

A cubic equation is a polynomial equation of degree 3, such as x³ − 6x² + 11x − 6 = 0. A quadratic has at most two roots. A cubic has at most three. A higher-degree equation (degree 4 or more) follows the same idea: degree n gives at most n roots.

There is no short formula you can use in the exam, as there is for a quadratic. Instead you find one root by trial, then reduce the cubic to a quadratic. That is the whole strategy.

The factor theorem makes this work. If f(a) = 0, then (x − a) divides f(x) exactly. So a root you find by trial gives you a factor. Dividing f(x) by that factor leaves a quadratic, and you already know how to solve that.

Which values should you try? For an equation with integer coefficients, a whole-number root must divide the constant term. For x³ − 6x² + 11x − 6, the constant is −6, so try ±1, ±2, ±3, ±6. This turns a guessing game into a short list.

Synthetic division is the fast way to divide by (x − a). You work only with the coefficients, so you avoid writing powers of x. It also shows the remainder, which equals f(a). A zero remainder confirms a root.

Key formulas to remember

Factor theorem: (x − a) is a factor of f(x) ⇔ f(a) = 0; If f(a) = 0, then a is a root. If the remainder on dividing by (x − a) is not 0, then a is not a root.
Remainder theorem: Remainder when f(x) is divided by (x − a) = f(a); Synthetic division gives this remainder as its last number.
Trial roots for integer coefficients: Any integer root divides the constant term. For a rational root p/q in lowest terms, p divides the constant term and q divides the leading coefficient.; Use this to build your list of values to test. A root need not be an integer if the leading coefficient is not 1.
Roots of ax³ + bx² + cx + d = 0 (a ≠ 0): α + β + γ = −b ÷ a; αβ + βγ + γα = c ÷ a; αβγ = −d ÷ a; Use these to check your roots or to eliminate options quickly.
Quick root checks: If a + b + c + d = 0, then x = 1 is a root. If −a + b − c + d = 0, then x = −1 is a root.; These are just f(1) = 0 and f(−1) = 0 for a cubic.
Number of roots: A polynomial equation of degree n has at most n roots; A cubic with real coefficients always has at least one real root. The other two may be real or non-real.

How to solve Cubic and Higher-Degree Equations questions

Use this method for any cubic question where the roots are numbers you can find by trial. It also works for degree 4 and above, one root at a time.

1Write the equation as f(x) = 0 with all terms on the left, in descending powers of x. If a power is missing, its coefficient is 0.
2List the possible trial roots: the factors of the constant term, both positive and negative. If the leading coefficient is not 1, also include fractions such as constant factor ÷ leading-coefficient factor.
3Test the easy values first: x = 1, then −1, then 2, −2, and so on. Stop at the first value where f(a) = 0.
4Divide f(x) by (x − a) using synthetic division. Write the coefficients in a row, bring down the first one, then multiply by a and add down each column. The last number must be 0.
5Read the quotient from the other numbers. It is a quadratic whose first coefficient is the cubic's first coefficient.
6Solve the quadratic by factorising or by the quadratic formula. The quadratic's roots are the other two roots of the cubic.
7Write all three roots. Check using the sum of roots (−b ÷ a) and the product of roots (−d ÷ a).

Quickest way: Substitute the options and use the sum of roots

When to use it: Use this when the question is an MCQ with roots or a factor listed in the options. It is usually faster than a full solution.

Check the quick tests first: a + b + c + d = 0 means 1 is a root. −a + b − c + d = 0 means −1 is a root.
Compute the sum of roots (−b ÷ a) and the product of roots (−d ÷ a). Cross out options whose roots do not match them.
If more than one option is left, substitute one root from each option into f(x). A single non-zero result removes that option.
For questions asking for an unknown constant k, substitute the given root into the equation and solve the resulting linear equation for k.
If the numbers get ugly after two quick tries, skip and return later. Each wrong answer costs 0.25 marks.

Common mistakes in Cubic and Higher-Degree Equations

Testing only positive values such as 1, 2, 3 and missing a negative root.
Students assume roots are positive because the equation looks like a positive-looking sum.
Fix: Always test both a and −a for each factor of the constant term. Use the check −a + b − c + d = 0 for x = −1.
Writing the factor as (x + a) when the root is a, or the reverse.
The sign flips between the root and the factor.
Fix: Root a gives factor (x − a). Root −2 gives factor (x + 2). In synthetic division, you always use the root itself as the multiplier.
Leaving out a missing term in synthetic division, for example x³ − 7x + 6 written as 1, −7, 6.
The x² term is not visible, so it is forgotten.
Fix: Write the coefficient 0 for any missing power. Here the coefficients are 1, 0, −7, 6.
Stopping after finding one root.
The first success feels like the answer.
Fix: A cubic has up to three roots. After the first root, always divide and solve the quadratic.
Getting the sign wrong in the sum or product of roots.
Students remember b ÷ a and d ÷ a without the minus signs.
Fix: Sum of roots = −b ÷ a. Product of roots = −d ÷ a. Product of pairs = c ÷ a (no minus sign). Test with a simple cubic like (x−1)(x−2)(x−3).
Assuming a non-zero remainder is fine and carrying on with the quotient.
Arithmetic slips are not caught, or the value tried is not actually a root.
Fix: The last number in synthetic division must be 0. If not, recheck the arithmetic once, then try another value.

Worked examples

Example 1

The roots of x³ − 6x² + 11x − 6 = 0 are: (a) 1, 2, 3 (b) −1, 2, 3 (c) 1, −2, 3 (d) 1, 2, −3

Show the solution

Add the coefficients: 1 − 6 + 11 − 6 = 0. So x = 1 is a root and (x − 1) is a factor.
Synthetic division by 1 on coefficients 1, −6, 11, −6: bring down 1. Next: −6 + 1 = −5. Next: 11 + (−5) = 6. Last: −6 + 6 = 0. The remainder is 0.
The quotient is x² − 5x + 6 = (x − 2)(x − 3), so the other roots are 2 and 3.
Check: sum of roots = 1 + 2 + 3 = 6 = −(−6) ÷ 1. Product = 6 = −(−6) ÷ 1.

Answer: (a) 1, 2, 3

Example 2

Which of the following is NOT a root of 2x³ − 3x² − 11x + 6 = 0? (a) 3 (b) 1/2 (c) −2 (d) 2

Show the solution

Test x = 3: 2(27) − 3(9) − 11(3) + 6 = 54 − 27 − 33 + 6 = 0. So 3 is a root.
Synthetic division by 3 on coefficients 2, −3, −11, 6: bring down 2. Next: −3 + 6 = 3. Next: −11 + 9 = −2. Last: 6 + (−6) = 0.
The quotient is 2x² + 3x − 2 = (2x − 1)(x + 2). The other roots are 1/2 and −2.
So the roots are 3, 1/2 and −2. Check: sum = 3 + 1/2 − 2 = 3/2 = −(−3) ÷ 2. ✓
Test x = 2 directly: 2(8) − 3(4) − 11(2) + 6 = 16 − 12 − 22 + 6 = −12, which is not 0. So 2 is not a root.

Answer: (d) 2

Example 3

If x = 3 is a root of x³ − 2x² − 5x + k = 0, the value of k is: (a) −6 (b) 6 (c) 3 (d) 12

Show the solution

Since 3 is a root, substitute x = 3 into the equation.
27 − 2(9) − 5(3) + k = 0, so 27 − 18 − 15 + k = 0.
This gives −6 + k = 0, so k = 6.
Check: with k = 6, x = 1 gives 1 − 2 − 5 + 6 = 0, so 1 is also a root. Dividing by (x − 1) leaves x² − x − 6 = (x − 3)(x + 2). The roots are 1, 3 and −2, which includes 3. ✓

Answer: (b) 6

Exam tips

Questions are usually built so that x = 1, −1, 2 or −2 is a root. Try those four first before building a long list.
With four options, check the sum of roots (−b ÷ a) first. It removes most wrong options in under 20 seconds.
For questions with an unknown constant, substitute the given root. Do not divide polynomials; that is slower and invites errors.
Do the quick check a + b + c + d = 0 before anything else. It costs five seconds and often gives you the first root.
If your trial list runs past six values with no root, skip the question. The negative marking means a guess is not free.

Practice questions from Equations

Cubic and Higher-Degree Equations: frequently asked questions

What is the difference between the roots of a quadratic and a cubic equation?

A quadratic has at most two roots, and a cubic has at most three. A quadratic can be solved with a formula. A cubic is usually solved by finding one root by trial and reducing it to a quadratic. A cubic with real coefficients always has at least one real root.

How do I solve a cubic equation using the factor theorem?

Find a value a where f(a) = 0 by testing factors of the constant term. Then (x − a) is a factor. Divide f(x) by (x − a) to get a quadratic, and solve it to find the other two roots.

Which values should I try first as roots?

Try 1 and −1 first, then 2 and −2. Use the factors of the constant term as your list. If the leading coefficient is not 1, also try fractions with that coefficient's factors in the denominator.

Do I need synthetic division, or is long division enough?

Both give the same answer. Synthetic division is shorter because it uses coefficients only, so you write fewer symbols and make fewer slips. Just remember to write 0 for any missing power.

Can a cubic equation have only one real root?

Yes. It can have one real root and two non-real roots. In CA Foundation questions, the equations are usually chosen so that all roots are real numbers you can find by trial.