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The remainder and factor theorems turn long division of polynomials into a single substitution, and they let you factorise cubics that no identity will crack. This post explains both, works three examples, and sets out exactly where they sit in the current CBSE syllabus, because that changed this year.
Are the remainder and factor theorems still in the Class 9 syllabus?
Not in CBSE's chapter list for 2026-27. The CBSE Mathematics curriculum for 2025-26 listed, under Class IX Polynomials, "Motivate and State the Remainder Theorem with examples" and "Statement and proof of the Factor Theorem", with factorisation of cubic polynomials using it. The 2026-27 Class IX curriculum has a chapter called Introduction to Polynomials whose key concepts are algebraic expressions, the definition and degree of a polynomial, linear polynomials, linear growth and decay, and the slope and y-intercept of y = ax + b. Factorisation moves to a separate chapter, Exploring Algebraic Identities. The same document does list "States and proves remainder theorem, factor theorem, and division algorithm" among the competencies for the secondary stage, but not in the Class 9 chapter content. The 2026-27 Class X curriculum lists zeros of a polynomial and the relation between zeros and coefficients of quadratics.
So: if your school follows the 2026-27 CBSE syllabus, ask your teacher whether these theorems will be assessed. If you are on another board, using the earlier books, or simply want a tool for factorising cubics, read on. NCERT's new textbook covers the linear polynomial side in Chapter 2, Introduction to Linear Polynomials.
What does the remainder theorem say, and why is it true?
Divide any polynomial p(x) by (x minus a). You get a quotient q(x) and a remainder r, which is a number because the divisor has degree 1:
p(x) = (x minus a) q(x) + r
Now put x = a. The first term becomes zero, so p(a) = r. That is the whole proof: the remainder on dividing by (x minus a) is p(a).
Worked example 1: find the remainder when p(x) = x³ + 3x² − 5x + 4 is divided by x − 2.
Substitute x = 2: 8 + 12 − 10 + 4 = 14. The remainder is 14.
Check by long division: the quotient is x² + 5x + 5, and (x − 2)(x² + 5x + 5) = x³ + 3x² − 5x − 10. Adding 14 gives back p(x).
For a divisor like 2x − 1, substitute the value that makes it zero, x = 1/2: p(1/2) = 1/8 + 3/4 − 5/2 + 4 = 19/8.
How do you use the factor theorem to factorise a cubic?
The factor theorem is the special case r = 0: (x minus a) is a factor of p(x) exactly when p(a) = 0.
Worked example 2: factorise p(x) = x³ − 6x² + 11x − 6.
Try small integers that divide the constant term 6: ±1, ±2, ±3, ±6.
p(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor.
Divide: x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6).
Factorise the quadratic: x² − 5x + 6 = (x − 2)(x − 3).
Answer: (x − 1)(x − 2)(x − 3).
Check by expanding (x − 1)(x² − 5x + 6) = x³ − 5x² + 6x − x² + 5x − 6 = x³ − 6x² + 11x − 6.
Worked example 3: find k if (x − 1) is a factor of 2x³ + kx² − 3x + 4.
p(1) = 2 + k − 3 + 4 = k + 3. Set this to zero: k = −3.
Check: 2 − 3 − 3 + 4 = 0.
What mistakes cost marks here?
Wrong sign on substitution. For x + 3 you substitute x = −3, not 3. Write "x + 3 = 0 gives x = −3" as a line of working.
Forgetting the remainder is a number. Dividing by a linear divisor never leaves an x in the remainder. If yours does, the division went wrong.
Stopping after one factor. Finding (x − 1) is step one. You still need to divide and factorise the quadratic.
Skipping the check. Expanding your factors takes a minute and shows the examiner, and you, that the answer is right.
How does this link to the new linear polynomial work?
A linear polynomial ax + b has one zero, at x = −b/a, and its graph y = ax + b is a straight line with slope a that crosses the x-axis there. The factor theorem says the same thing for bigger polynomials: every zero a gives a factor (x − a), and on a graph it is a point where the curve meets the x-axis. Our post on linear equations in Class 9 covers the straight-line side, and polynomials in Class 10 picks up zeros and coefficients.
A practice routine for polynomials
Mixing problem types makes you choose the method yourself, which is what an exam asks. Rohrer and Taylor (2007) found that shuffling mathematics problems improved learning compared with practising one type at a time.
Make a set of eight: three remainders, three factorisations, two "find k" questions. Shuffle them.
Before each, write which theorem you are using and what you will substitute.
Check every answer by expanding or by long division.
Repeat with a fresh set three days later.
If you keep landing on the wrong factor, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see which step went wrong.
Start by checking which polynomial topics your school's 2026-27 syllabus lists, then try worked example 2 without looking.
