Types of polynomials — by degree and terms, with factoring strategies

easy CBSE 3 min read

Question

Classify 3x42x2+73x^4 - 2x^2 + 7 by degree and number of terms. What factoring strategy works for each polynomial type?

(CBSE 9 & 10 — polynomials chapter)


Solution — Step by Step

DegreeNameExample
0Constant77
1Linear3x+23x + 2
2Quadraticx25x+6x^2 - 5x + 6
3Cubicx3+1x^3 + 1
4Quartic (Biquadratic)3x42x2+73x^4 - 2x^2 + 7

Our polynomial has degree 4 (highest power of xx), so it’s a quartic/biquadratic polynomial.

TermsName
1Monomial (e.g., 5x35x^3)
2Binomial (e.g., x24x^2 - 4)
3Trinomial (e.g., x2+3x+2x^2 + 3x + 2)

3x42x2+73x^4 - 2x^2 + 7 has 3 terms → trinomial.

TypeStrategy
Common factorTake out GCF first (always check this!)
Quadratic trinomialSplit middle term or use formula
Difference of squaresa2b2=(a+b)(ab)a^2 - b^2 = (a+b)(a-b)
Perfect squarea2±2ab+b2=(a±b)2a^2 \pm 2ab + b^2 = (a \pm b)^2
Sum/difference of cubesa3±b3=(a±b)(a2ab+b2)a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)
BiquadraticSubstitute t=x2t = x^2, reduce to quadratic

Why This Works

graph TD
    A["Factor a polynomial"] --> B{"Common factor?"}
    B -->|"Yes"| C["Factor out GCF first"]
    B -->|"No"| D{"Degree?"}
    C --> D
    D -->|"2 (quadratic)"| E{"Type?"}
    D -->|"3 (cubic)"| F["Try factor theorem<br/>or grouping"]
    D -->|"4 (biquadratic)"| G["Substitute t = x²<br/>Solve as quadratic in t"]
    E -->|"a² - b²"| H["(a+b)(a-b)"]
    E -->|"Trinomial"| I["Split middle term<br/>or quadratic formula"]
    E -->|"Perfect square"| J["(a ± b)²"]

Classification guides strategy. A quadratic trinomial calls for middle-term splitting. A biquadratic with only even powers calls for substitution. Knowing the type saves time by pointing you to the right technique immediately.


Alternative Method — Factor Theorem

For any polynomial p(x)p(x): if p(a)=0p(a) = 0, then (xa)(x - a) is a factor. Try small integer values (0,±1,±20, \pm 1, \pm 2) to find roots. This works for all degrees but is most practical for cubics and quartics.

For CBSE 10: the factor theorem combined with synthetic division is the fastest way to factor cubics. Find one root by trial, divide out the linear factor, and you’re left with a quadratic that you can factor normally.


Common Mistake

Students try to factor x42x2+7x^4 - 2x^2 + 7 by splitting the middle term as if it were a quadratic in xx. But it’s a quadratic in x2x^2 — substitute t=x2t = x^2 first to get 3t22t+73t^2 - 2t + 7, then check if this quadratic factors. Here the discriminant is 484=80<04 - 84 = -80 < 0, so it doesn’t factor over the reals. Not every polynomial can be factored — recognising this saves time.

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