A polynomial is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where the exponents are non-negative integers and the coefficients are real numbers. That is it — no x⁻², no √x, no 1/x lurking in there.
The degree of a polynomial is the highest power with a non-zero coefficient. A degree-1 polynomial is linear, degree-2 is quadratic, degree-3 is cubic, and degree-4 is biquadratic. The degree matters because it tells you the maximum number of zeroes a polynomial can have.
A zero (or root) of a polynomial p(x) is a value c such that p(c) = 0. Think of it as the x-coordinate where the polynomial's graph touches or crosses the x-axis. A degree-n polynomial has exactly n zeroes, counting multiplicity, in the complex number system — though for CDS, you only deal with real zeroes.
Here is a useful analogy. Think of a polynomial as a number written in base x instead of base 10. Just as the number 3452 breaks into 3×10³ + 4×10² + 5×10 + 2, a cubic polynomial 3x³ + 4x² + 5x + 2 is structurally the same — a sum of powers with coefficients. Operations on polynomials (addition, subtraction, multiplication, division) follow the same logic as arithmetic on whole numbers. This is why the Remainder Theorem and Factor Theorem are really just the polynomial versions of long division with a remainder.
For CDS, polynomials appear in two flavours of question. First, straightforward application of Remainder/Factor theorems where you substitute a value and solve for an unknown coefficient. Second, abstract questions about the relationship between zeroes and coefficients (Vieta's formulas). Both are very doable once you know what each theorem actually says.
When you divide a polynomial p(x) by a linear factor (x − c), the remainder is exactly p(c).
Why does this work? Polynomial long division gives you: p(x) = (x − c) · q(x) + R, where R is a constant (the remainder, since the divisor is degree 1). Substituting x = c: p(c) = (c − c) · q(c) + R = R. So the remainder equals p(c). No division needed — just substitute.
Practical use: If a question asks "what is the remainder when p(x) is divided by (x − 3)?", compute p(3) in five seconds. Done.
The Factor Theorem is the Remainder Theorem with one extra condition: (x − c) is a factor of p(x) if and only if p(c) = 0. This is just the observation that remainder zero means exact divisibility.
Use case: to verify that a given value is a zero of a polynomial, substitute it. To find the remaining factors after confirming one zero, perform synthetic division or polynomial long division.
For a monic polynomial (leading coefficient = 1) of degree n with zeroes α₁, α₂, ..., αₙ:
= −(coefficient of xⁿ⁻¹)= +(coefficient of xⁿ⁻²)= −(coefficient of xⁿ⁻³)= (−1)ⁿ × (constant term)For a cubic x³ + px² + qx + r with zeroes α, β, γ:
α + β + γ = −pαβ + βγ + γα = qαβγ = −rFor a biquadratic x⁴ + ax³ + bx² + cx + d with zeroes α, β, γ, δ:
α + β + γ + δ = −aαβ + αγ + αδ + βγ + βδ + γδ = bαβγ + αβδ + αγδ + βγδ = −cαβγδ = dThese formulas let you find relationships between zeroes without factoring the polynomial completely.
When one zero c is known, divide p(x) by (x − c) using synthetic division to get a degree (n−1) polynomial. Repeat until you reach a quadratic — then apply the sum-product method or the quadratic formula.
Synthetic division (for dividing by x − c): write only the coefficients, bring down the first, multiply by c, add down the column. This is faster than long division for CDS purposes.
Some CDS questions exploit special factoring identities:
xⁿ − 1 is always divisible by (x − 1).xⁿ + 1 is divisible by (x + 1) only when n is odd.x² + 1 divides x⁴ − 1 since x⁴ − 1 = (x² − 1)(x² + 1).1 + x² + x⁴ + x⁶ = (1 + x²)(1 + x⁴) — this factoring came up directly in a 2024 CDS question.When a polynomial in x has a geometric-series structure (consecutive even or odd powers), factor out the common term first, then recognise the sum-of-geometric-series pattern.
When asked for the remainder of p(x) ÷ (x − c), skip division entirely. Just compute p(c).
Example: remainder of x³ − 4x² + 6x − 2 divided by (x − 2).
Standard long division: ~45 seconds.
Substitution: p(2) = 8 − 16 + 12 − 2 = 2. Done in 8 seconds.
The step count drops from roughly 12 steps (long division) to 4 steps (substitute, arithmetic, answer).
Questions ask for α² + β² + γ² for a cubic. Don't find each zero individually.
Use the identity: α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα).
For the cubic x³ + px² + qx + r: this equals p² − 2q.
This saves you from solving a cubic. The substitution approach for this pattern takes 3 steps; factoring the cubic from scratch can take 8-10 steps.
When a polynomial looks like xᵃ + xᵃ⁺² + xᵃ⁺⁴ + xᵃ⁺⁶, factor out xᵃ:
xᵃ(1 + x² + x⁴ + x⁶)
Then factor the bracket as a geometric series sum: (1 + x²)(1 + x⁴).
This immediately tells you the polynomial is divisible by both (x² + 1) and (x⁴ + 1) — no trial division needed. Standard divisibility-check by substitution would require testing ±i for statement 1 and checking the four fourth roots of −1 for statement 2. The factoring route cuts this to a single algebraic manipulation (~10 seconds vs ~90 seconds).
When one zero c is given and you need properties of the other zeroes:
Step 1 — Find the unknown coefficient using p(c) = 0 (Remainder Theorem).
Step 2 — Divide p(x) by (x − c) using synthetic division to get the depressed polynomial.
Step 3 — Apply Vieta's or factorise the depressed polynomial.
This pipeline appeared directly in CDS 2024. Executing it cold in the exam takes about 90 seconds. Without the pipeline (guessing-and-checking zeroes) can take 3-4 minutes.
Students constantly get the sign of αβγ wrong for a cubic x³ + px² + qx + r. The pattern:
Sum of zeroes: −p (minus the coefficient of x²)
Product of zeroes: −r (minus the constant term, since n=3 is odd)
Mnemonic: for odd-degree polynomials, the product of all zeroes carries a negative sign on the constant. For even-degree (biquadratic), the product carries a positive sign. One check: for x² − 5x + 6 = (x−2)(x−3), product = 6 = positive constant. Degree is even. This sign check takes 2 seconds and prevents a full sign-error in Vieta's.
When a polynomial question appears, classify it in the first 10 seconds:
Type A — "Find the remainder" → Use the Remainder Theorem. Substitute x = c directly. No division.
Type B — "Is (x − c) a factor?" or "Find the unknown coefficient given a zero" → Use the Factor Theorem. Set p(c) = 0, solve for the unknown.
Type C — "Sum/product/sum-of-squares of zeroes" → Use Vieta's formulas. Do not find individual zeroes unless the polynomial is already depressed to a quadratic.
Type D — "One zero given, find properties of others" → Pipeline: Factor Theorem to find unknown coefficient → synthetic division → Vieta's or factoring on the depressed polynomial.
Type E — "Is this polynomial divisible by that one?" → Check if the zeroes of the divisor are also zeroes of the dividend, or look for a geometric-series / algebraic factoring.
Decision rule: never apply the quadratic formula to a depressed quadratic if sum-product factoring takes under 10 seconds. Save the formula for irrationals.
Why this question: This is the canonical Type D question from CDS 2024 — given one zero, find a coefficient, then get properties of remaining zeroes. The pipeline is almost automatic once you have practised it.
Solving path:
p(2) = 0 gives 8 + 12 − 12 − a = 0, so a = 8.p(x) = x³ + 3x² − 6x − 8. Divide by (x − 2) via synthetic division: coefficients 1, 3, −6, −8 divided by root 2 give 1, 5, 4, 0. Depressed polynomial: x² + 5x + 4.(x + 1)(x + 4). Other zeroes are −1 and −4.(−1)² + (−4)² = 1 + 16 = 17. Answer: option (b).Note: the answer key correctly identifies option (b) as 17, not option (c) as 21. Check your option letters carefully in the exam.
Why this question: This abstract question tests whether you understand that a monic degree-4 polynomial is completely determined by its four zeroes. The algebra here is purely about factored form — no numerical substitution.
Solving path:
p(x) = (x − α)(x − β)(x − γ)(x − δ) and q(x) = (x − α)(x − β)(x − γ)(x − λ).p(x) − q(x) = (x − α)(x − β)(x − γ) [(x − δ) − (x − λ)].(x − δ) − (x − λ) = −δ + λ = λ − δ.(x − α)(x − β)(x − γ): result is λ − δ.The answer is λ − δ, which corresponds to option (b). The key move is writing both polynomials in factored form — do not expand and compare coefficients, that wastes two minutes.
Why this question: This tests geometric-series factoring and divisibility. The polynomial structure x^(4k) + x^(4k+2) + x^(4k+4) + x^(4k+6) looks intimidating, but factoring reveals everything in one line.
Solving path:
x^(4k): the expression becomes x^(4k)(1 + x² + x⁴ + x⁶).x², four terms. Factor it: (1 + x²)(1 + x⁴).x^(4k)(1 + x²)(1 + x⁴).x² + 1? Yes — (1 + x²) is a factor. Remainder = 0. Correct.x⁴ + 1? Yes — (1 + x⁴) is a factor. Remainder = 0. Correct.The official answer key confirms option (c). The shortcut is the factoring of 1 + x² + x⁴ + x⁶ — once you see it is (1 + x²)(1 + x⁴), both statements resolve simultaneously.
Confusing the sign in the Remainder Theorem when the divisor is (x + c). The divisor (x + c) equals (x − (−c)), so you substitute x = −c, not x = c. Students who write p(c) for a (x + c) divisor lose the mark every time.
Using Vieta's formulas on a non-monic polynomial without normalising. Vieta's formulas in the standard form assume the leading coefficient is 1. For 2x³ + 6x² − 12x − 16, you must first divide through by 2 to get x³ + 3x² − 6x − 8 before applying sum = −(coefficient of x²) = −3.
Assuming a degree-3 polynomial has three distinct real zeroes. Some cubics have one real zero and two complex conjugate zeroes. On CDS, this typically means only one rational zero exists — use the Rational Root Theorem or synthetic division, not a guess that all three are clean integers.
Forgetting to check the sign of the product of zeroes. For cubic x³ + px² + qx + r, the product αβγ = −r. Students frequently write +r. The sign alternates with degree: positive for degree 2, negative for degree 3, positive for degree 4.
Attempting to divide p(x) − q(x) by algebraically expanding both degree-4 polynomials. In the abstract question type (PYQ id 6a206e02ccfcbabc57a3ad9b), expanding gives you a₃ − b₃, a₂ − b₂, etc. as coefficients — you cannot simplify further without knowing the individual coefficients. The factored-form approach is the only clean route.
Misidentifying the geometric-series factoring. 1 + x² + x⁴ + x⁶ is a 4-term geometric series. Students sometimes mistake it for a sum of squares (1 + x²)² = 1 + 2x² + x⁴ — which it is not. The correct factoring is (1 + x²)(1 + x⁴), verifiable by expanding: 1 + x⁴ + x² + x⁶.