Polynomial Theorems and Roots

Fundamental Theorem of Algebra and finding roots

Polynomial Theorems

Fundamental Theorem of Algebra

Every polynomial of degree nn has exactly nn roots (counting multiplicity), including complex roots.

Example: A degree 3 polynomial has 3 roots (some may be repeated or complex).

Rational Root Theorem

If pq\frac{p}{q} is a rational root of anxn+...+a0=0a_nx^n + ... + a_0 = 0, then:

  • pp divides a0a_0 (constant term)
  • qq divides ana_n (leading coefficient)

Example: For 2x3+5x2−4x−3=02x^3 + 5x^2 - 4x - 3 = 0

Possible rational roots: ±1,±3,±12,±32\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}

Descartes' Rule of Signs

Count sign changes in P(x)P(x) to find:

  • Positive roots: Number of sign changes (or less by an even number)
  • Negative roots: Number of sign changes in P(−x)P(-x)

Complex Conjugate Theorem

If a+bia + bi is a root of a polynomial with real coefficients, then a−bia - bi is also a root.

Example: If 2+3i2 + 3i is a root, then 2−3i2 - 3i must also be a root.

📚 Practice Problems

1Problem 1easy

❓ Question:

List all possible rational roots: x3−5x2+2x−6=0x^3 - 5x^2 + 2x - 6 = 0

💡 Show Solution

Use the Rational Root Theorem:

Factors of constant term (−6-6): ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6 Factors of leading coefficient (11): ±1\pm 1

Possible rational roots: ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6

Answer: ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6

2Problem 2medium

❓ Question:

A polynomial has roots 22, −3-3, and 1+i1 + i. What is the minimum degree?

💡 Show Solution

We have roots: 22, −3-3, and 1+i1 + i

By the Complex Conjugate Theorem, if 1+i1 + i is a root, then 1−i1 - i must also be a root (assuming real coefficients).

Total roots: 22, −3-3, 1+i1 + i, 1−i1 - i

That's 4 roots, so minimum degree is 4.

Answer: Degree 4

3Problem 3hard

❓ Question:

Find a polynomial with real coefficients that has roots 33 and 2−i2 - i

💡 Show Solution

Since coefficients are real and 2−i2 - i is a root, then 2+i2 + i must also be a root.

Roots: 33, 2−i2 - i, 2+i2 + i

Step 1: Write factors (x−3)(x−(2−i))(x−(2+i))(x - 3)(x - (2 - i))(x - (2 + i))

Step 2: Multiply the complex factors first (x−2+i)(x−2−i)(x - 2 + i)(x - 2 - i) =[(x−2)+i][(x−2)−i]= [(x - 2) + i][(x - 2) - i] =(x−2)2−i2= (x - 2)^2 - i^2 =(x−2)2+1= (x - 2)^2 + 1 =x2−4x+4+1= x^2 - 4x + 4 + 1 =x2−4x+5= x^2 - 4x + 5

Step 3: Multiply by (x−3)(x - 3) (x−3)(x2−4x+5)(x - 3)(x^2 - 4x + 5) =x3−4x2+5x−3x2+12x−15= x^3 - 4x^2 + 5x - 3x^2 + 12x - 15 =x3−7x2+17x−15= x^3 - 7x^2 + 17x - 15

Answer: P(x)=x3−7x2+17x−15P(x) = x^3 - 7x^2 + 17x - 15

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