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CDS Mathematics · Polynomials

Degree, Zeros and Coefficients

Degrees add when polynomials are multiplied and can drop when they are added; the zeros of a polynomial are tied to its coefficients by sums and products.

All Polynomials notesCDS Mathematics strategy

Why this matters

Eight PYQs. Two facts about degree, one about identities (every coefficient must vanish), and the sum-and-product relations for the zeros — the same ones as for a quadratic, extended to a cubic.

Concept 1 of 2: Degree, identities and integer values

Multiplying leading terms can never cancel, so degrees add. Adding two polynomials of the same degree can cancel the leading terms, so the degree can fall. A polynomial that is zero for EVERY x must have every coefficient zero.

Definition

  • deg⁡(fg)=deg⁡f+deg⁡g\deg(fg) = \deg f + \deg g.
  • deg⁡(f±g)≤max⁡(deg⁡f,deg⁡g)\deg(f \pm g) \le \max(\deg f, \deg g), with equality when the degrees differ.
  • ax2+bx+c=0ax^2 + bx + c = 0 for all xx (an identity) only when a=b=c=0a = b = c = 0.
  • A polynomial can take integer values at every integer without having integer coefficients: x(x+1)2\dfrac{x(x + 1)}{2} is always an integer. The constant term, the value at 00, must be an integer.

Degree of a product

deg⁡(fg)=deg⁡f+deg⁡g\deg(fg) = \deg f + \deg g

Worked example

For which kk is (k2−1)x2+(k2+k)x+(k+1)=0(k^2 - 1)x^2 + (k^2 + k)x + (k + 1) = 0 an identity?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (I) 2026 — Elementary Mathematics · Q2Moderate

Example 1 · Polynomials · Degree, Zeros and Coefficients

What is the value of k for which (k2−5k+4)x2+(k2−3k−4)x+(k2−4k)=0(k^2 - 5k + 4)x^2 + (k^2 - 3k - 4)x + (k^2 - 4k) = 0 is an identity ?

Equal degrees can cancel

(x3+x)−(x3+1)=x−1(x^3 + x) - (x^3 + 1) = x - 1: the degree dropped from 33 to 11. So the degree of a sum is only 'at most' the larger degree.
Drill 3 more on degree, identities and integer values

Concept 2 of 2: Zeros and coefficients

A monic polynomial with zeros α,β,γ\alpha, \beta, \gamma is (x−α)(x−β)(x−γ)(x - \alpha)(x - \beta)(x - \gamma). Expanding it shows the coefficients are sums and products of the zeros, exactly as for a quadratic.

Definition

For ax3+bx2+cx+dax^3 + bx^2 + cx + d with zeros α,β,γ\alpha, \beta, \gamma:

  • α+β+γ=−ba\alpha + \beta + \gamma = -\dfrac ba, αβ+βγ+γα=ca\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac ca, αβγ=−da\alpha\beta\gamma = -\dfrac da;
  • α2+β2+γ2=(ba)2−2ca\alpha^2 + \beta^2 + \gamma^2 = \left(\dfrac ba\right)^2 - \dfrac{2c}{a}.
  • A polynomial with new zeros: find their sum and product, then write x2−(sum)x+productx^2 - (\text{sum})x + \text{product} and clear fractions.
  • Two monic polynomials sharing three zeros differ by (common cubic) ×\times (difference of the fourth factors).

Cubic: sum of zeros

α+β+γ=−ba,αβ+βγ+γα=ca,αβγ=−da\alpha + \beta + \gamma = -\dfrac ba, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \dfrac ca, \quad \alpha\beta\gamma = -\dfrac da

Worked example

The zeros of 2x3−3x2−3x+22x^3 - 3x^2 - 3x + 2 are α,β,γ\alpha, \beta, \gamma. Find 1α+1β+1γ\dfrac1\alpha + \dfrac1\beta + \dfrac1\gamma.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q50Moderate

Example 2 · Polynomials · Degree, Zeros and Coefficients

If α\alpha, β\beta and γ\gamma are the zeros of the polynomial f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d, then α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 is equal to

Divide by a

The relations use ba\dfrac ba, ca\dfrac ca, da\dfrac da. For a non-monic cubic, ∑α2=b2−2aca2\sum\alpha^2 = \dfrac{b^2 - 2ac}{a^2}, not b2−2cb^2 - 2c.
Drill 3 more on zeros and coefficients

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Degree, identities and integer values

    Degree of a product

    deg⁡(fg)=deg⁡f+deg⁡g\deg(fg) = \deg f + \deg g
  • Zeros and coefficients

    Cubic: sum of zeros

    α+β+γ=−ba,αβ+βγ+γα=ca,αβγ=−da\alpha + \beta + \gamma = -\dfrac ba, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \dfrac ca, \quad \alpha\beta\gamma = -\dfrac da

Watch out for (2)

Test yourself on Polynomials

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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