Học · Algebra 2

Polynomial Functions: End Behavior, Roots, and the Rational Root Theorem

The Fundamental Theorem of Algebra guarantees a polynomial of degree n has n roots. Master the Rational Root Theorem, end-behavior rules, and you'll factor polynomials too big for plain trial and error.

10 phút TEKS 7A,7C,7D,7E,7F,7I Algebra 2

Higher degrees, same logic

Algebra 1 factored quadratics. Algebra 2 takes you up to degree 5 and beyond. The Fundamental Theorem of Algebra says any degree-n polynomial has exactly n roots (counting multiplicity, including complex roots).

End behavior depends on degree and leading coefficient

Even degree, positive leading: both ends → +∞ Even degree, negative leading: both ends → −∞ Odd degree, positive leading: left → −∞, right → +∞ Odd degree, negative leading: left → +∞, right → −∞
even, + even, − odd, + odd, −
Four end-behavior shapes. Sign of leading coefficient flips the right end; parity of degree decides whether both ends match.
Identify end behavior
What is the end behavior of f(x) = −2x⁴ + 3x² + 5?

A degree-n polynomial has up to n real roots

Fundamental Theorem of Algebra

A polynomial of degree n has exactly n roots (real + complex, counting multiplicity). At most n of them can be real.

Maximum real roots
A polynomial has a degree of 5. What is the maximum number of real roots it can have?

The Rational Root Theorem

For a polynomial with integer coefficients, every rational root has the form ±(factor of constant term)/(factor of leading coefficient).

x³ − 4x² + x + 6 = 0 Constant = 6 → factors ±1, ±2, ±3, ±6 Leading coefficient = 1 → factors ±1 Possible rational roots: ±1, ±2, ±3, ±6 Test these one by one (synthetic division). Once you find one root, divide and reduce to a quadratic.
List the candidates
According to the Rational Root Theorem, possible rational roots of x³ − 4x² + x + 6 = 0 must come from:

Synthetic division (the speed factor)

Synthetic division is a streamlined way to divide a polynomial by (x − r). If the remainder is 0, then r is a root and the quotient is the factor you want.

Divide x³ − 4x² + x + 6 by (x − 3): Coefficients: [1, −4, 1, 6], divisor 3 Bring down 1; 1·3 = 3 → −4 + 3 = −1 −1·3 = −3 → 1 + (−3) = −2 −2·3 = −6 → 6 + (−6) = 0  (remainder = 0, so 3 IS a root) Quotient: x² − x − 2 = (x − 2)(x + 1) All roots: x = 3, 2, −1

3-second recap

  • Degree n → at most n real roots, exactly n total (counting complex).
  • End behavior: even/odd × +/− gives 4 cases.
  • Rational Root Theorem: candidates = ±(const factors)/(leading factors).
  • Synthetic division: faster than long division; remainder 0 means you found a root.

Check yourself

Quick check #1
What is the end behavior of f(x) = −2x³ + 5?
Quick check #2
How many real roots does f(x) = (x − 1)(x + 2)(x − 5) have?