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Fundamental Theorem of Algebra and Factoring

Apply the Fundamental Theorem of Algebra to find all roots of polynomials and completely factor polynomial expressions.

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Fundamental Theorem of Algebra and Factoring

The Fundamental Theorem of Algebra

Theorem: Every polynomial equation of degree n≥1n \geq 1 with complex coefficients has exactly nn complex roots (counting multiplicity).

Implications

  1. A polynomial of degree nn has exactly nn roots in the complex number system
  2. Some roots may be repeated (multiplicity > 1)
  3. Some roots may be real, some may be complex
  4. Complex roots with real coefficients come in conjugate pairs

Example

The polynomial p(x)=x3−6x2+11x−6p(x) = x^3 - 6x^2 + 11x - 6 is degree 3, so it has exactly 3 roots.

If we factor: p(x)=(x−1)(x−2)(x−3)p(x) = (x - 1)(x - 2)(x - 3)

The three roots are x=1,2,3x = 1, 2, 3 (all real).

Relationship Between Roots and Factors

If rr is a root of p(x)p(x), then (x−r)(x - r) is a factor of p(x)p(x).

Conversely: If (x−r)(x - r) is a factor of p(x)p(x), then rr is a root of p(x)p(x).

Multiplicity

If (x−r)k(x - r)^k is a factor but (x−r)k+1(x - r)^{k+1} is not, then rr is a root of multiplicity kk.

Effects of multiplicity on graphs:

  • Multiplicity 1 (simple root): Graph crosses the x-axis
  • Multiplicity 2 (double root): Graph touches but doesn't cross (turning point)
  • Multiplicity 3: Graph crosses with a flattening
  • Even multiplicity: Graph touches x-axis
  • Odd multiplicity: Graph crosses x-axis

Complete Factorization

Every polynomial with real coefficients can be factored into:

  1. Linear factors: (x−r)(x - r) where rr is a real root
  2. Irreducible quadratic factors: ax2+bx+cax^2 + bx + c where b2−4ac<0b^2 - 4ac < 0

Linear Factorization Theorem

If p(x)p(x) is a polynomial of degree n≥1n \geq 1, then: p(x)=an(x−r1)(x−r2)⋯(x−rn)p(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n)

where ana_n is the leading coefficient and r1,r2,…,rnr_1, r_2, \ldots, r_n are the roots (possibly complex, possibly repeated).

Finding All Roots

Step-by-step process:

  1. Count the roots: Degree nn means nn roots total
  2. Find rational roots: Use Rational Root Theorem
  3. Factor out linear factors: Use synthetic division
  4. Reduce the polynomial: Continue until you have a quadratic or simpler
  5. Find remaining roots: Use quadratic formula if needed
  6. Include complex conjugates: If you find a+bia + bi, also include a−bia - bi

Complex Conjugate Root Theorem

Theorem: If a polynomial has real coefficients and a+bia + bi (where b≠0b \neq 0) is a root, then the complex conjugate a−bia - bi is also a root.

Consequence for Factoring

If a+bia + bi and a−bia - bi are roots, the corresponding quadratic factor is: [x−(a+bi)][x−(a−bi)]=x2−2ax+(a2+b2)[x - (a + bi)][x - (a - bi)] = x^2 - 2ax + (a^2 + b^2)

This is a real quadratic (no imaginary coefficients).

Writing Polynomials from Roots

Given roots, we can construct the polynomial:

Example: Roots are 2,3,−12, 3, -1

Factors: (x−2)(x−3)(x+1)(x - 2)(x - 3)(x + 1)

Polynomial: p(x)=(x−2)(x−3)(x+1)=x3−4x2+x+6p(x) = (x - 2)(x - 3)(x + 1) = x^3 - 4x^2 + x + 6

With complex roots: If roots are 1,2+i,2−i1, 2 + i, 2 - i:

p(x)=(x−1)[x−(2+i)][x−(2−i)]p(x) = (x - 1)[x - (2 + i)][x - (2 - i)] =(x−1)[x2−4x+5]= (x - 1)[x^2 - 4x + 5] =x3−5x2+9x−5= x^3 - 5x^2 + 9x - 5

Number of Real vs Complex Roots

For a polynomial of degree nn with real coefficients:

  • Complex roots come in conjugate pairs
  • If nn is odd, there is at least one real root
  • If nn is even, there may be no real roots

Example Scenarios (degree 4):

  • 4 real roots
  • 2 real, 2 complex (conjugate pair)
  • 0 real, 4 complex (two conjugate pairs)

Descartes' Rule of Signs

For positive roots: Count sign changes in p(x)p(x). The number of positive real roots is either equal to the number of sign changes or less by an even number.

For negative roots: Count sign changes in p(−x)p(-x).

Example

p(x)=x3−4x2+x+6p(x) = x^3 - 4x^2 + x + 6

Signs: +−+++ - + + → 2 sign changes Positive roots: 2 or 0

p(−x)=−x3−4x2−x+6p(-x) = -x^3 - 4x^2 - x + 6

Signs: −−−+- - - + → 1 sign change Negative roots: 1

📚 Practice Problems

1Problem 1easy

❓ Question:

Find all roots of p(x)=x3−7x2+14x−8p(x) = x^3 - 7x^2 + 14x - 8 and write the complete factorization.

💡 Show Solution

Solution:

Given: p(x)=x3−7x2+14x−8p(x) = x^3 - 7x^2 + 14x - 8

Step 1: Check for rational roots

By Rational Root Theorem, possible rational roots are: ±1,±2,±4,±8\pm 1, \pm 2, \pm 4, \pm 8

Test x=1x = 1: p(1)=1−7+14−8=0p(1) = 1 - 7 + 14 - 8 = 0 ✓

So x=1x = 1 is a root, and (x−1)(x - 1) is a factor.

Step 2: Divide by (x−1)(x - 1) using synthetic division

& 1 & -7 & 14 & -8 \\ 1 & & 1 & -6 & 8 \\ \hline & 1 & -6 & 8 & 0 \end{array}$$ Result: $p(x) = (x - 1)(x^2 - 6x + 8)$ **Step 3: Factor the quadratic** $$x^2 - 6x + 8 = (x - 2)(x - 4)$$ **Step 4: Complete factorization** $$p(x) = (x - 1)(x - 2)(x - 4)$$ **All roots:** $x = 1, 2, 4$ **Verification:** - Degree 3 polynomial → 3 roots ✓ - All roots are real ✓ - Expand: $(x - 1)(x - 2)(x - 4)$ - $(x - 1)(x^2 - 6x + 8) = x^3 - 6x^2 + 8x - x^2 + 6x - 8 = x^3 - 7x^2 + 14x - 8$ ✓

2Problem 2medium

❓ Question:

A polynomial of degree 4 has roots at x=2x = 2 (multiplicity 2) and x=−1±ix = -1 \pm i. Find the polynomial in standard form with leading coefficient 1.

💡 Show Solution

Solution:

Given information:

  • Degree 4
  • Root: x=2x = 2 with multiplicity 2
  • Roots: x=−1+ix = -1 + i and x=−1−ix = -1 - i (complex conjugate pair)

Step 1: Write factors

From x=2x = 2 (multiplicity 2): (x−2)2(x - 2)^2

From x=−1+ix = -1 + i: [x−(−1+i)]=(x+1−i)[x - (-1 + i)] = (x + 1 - i)

From x=−1−ix = -1 - i: [x−(−1−i)]=(x+1+i)[x - (-1 - i)] = (x + 1 + i)

Step 2: Combine complex conjugate factors

[x−(−1+i)][x−(−1−i)]=[(x+1)−i][(x+1)+i][x - (-1 + i)][x - (-1 - i)] = [(x + 1) - i][(x + 1) + i]

This is a difference of squares pattern: (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2

=(x+1)2−i2=(x+1)2−(−1)=(x+1)2+1= (x + 1)^2 - i^2 = (x + 1)^2 - (-1) = (x + 1)^2 + 1

=x2+2x+1+1=x2+2x+2= x^2 + 2x + 1 + 1 = x^2 + 2x + 2

Step 3: Form the complete polynomial

p(x)=(x−2)2(x2+2x+2)p(x) = (x - 2)^2(x^2 + 2x + 2)

Step 4: Expand

First: (x−2)2=x2−4x+4(x - 2)^2 = x^2 - 4x + 4

Then multiply: (x2−4x+4)(x2+2x+2)(x^2 - 4x + 4)(x^2 + 2x + 2)

=x4+2x3+2x2−4x3−8x2−8x+4x2+8x+8= x^4 + 2x^3 + 2x^2 - 4x^3 - 8x^2 - 8x + 4x^2 + 8x + 8

=x4−2x3−2x2+8= x^4 - 2x^3 - 2x^2 + 8

Answer: p(x)=x4−2x3−2x2+8p(x) = x^4 - 2x^3 - 2x^2 + 8

Verification:

  • Degree 4 ✓
  • Leading coefficient 1 ✓
  • Has 4 roots total (counting multiplicity) ✓

3Problem 3hard

❓ Question:

Find all roots of p(x)=x4+5x2+4p(x) = x^4 + 5x^2 + 4 and write the complete factorization over the complex numbers.

💡 Show Solution

Solution:

Given: p(x)=x4+5x2+4p(x) = x^4 + 5x^2 + 4

Step 1: Recognize this as quadratic in form

Let u=x2u = x^2, then: u2+5u+4=0u^2 + 5u + 4 = 0

Step 2: Factor the quadratic u2+5u+4=(u+1)(u+4)=0u^2 + 5u + 4 = (u + 1)(u + 4) = 0

So u=−1u = -1 or u=−4u = -4

Step 3: Solve for xx

Case 1: x2=−1x^2 = -1 x=±ix = \pm i

Case 2: x2=−4x^2 = -4 x=±2ix = \pm 2i

All roots: x=i,−i,2i,−2ix = i, -i, 2i, -2i

Step 4: Write complete factorization

p(x)=(x−i)(x+i)(x−2i)(x+2i)p(x) = (x - i)(x + i)(x - 2i)(x + 2i)

Or in terms of real quadratics:

(x−i)(x+i)=x2+1(x - i)(x + i) = x^2 + 1 (x−2i)(x+2i)=x2+4(x - 2i)(x + 2i) = x^2 + 4

p(x)=(x2+1)(x2+4)p(x) = (x^2 + 1)(x^2 + 4)

Answer:

  • Complex factorization: p(x)=(x−i)(x+i)(x−2i)(x+2i)p(x) = (x - i)(x + i)(x - 2i)(x + 2i)
  • Real factorization: p(x)=(x2+1)(x2+4)p(x) = (x^2 + 1)(x^2 + 4)
  • Roots: x=±i,±2ix = \pm i, \pm 2i

Verification:

  • Degree 4 → 4 roots ✓
  • All roots are purely imaginary (no real roots) ✓
  • Expand: (x2+1)(x2+4)=x4+4x2+x2+4=x4+5x2+4(x^2 + 1)(x^2 + 4) = x^4 + 4x^2 + x^2 + 4 = x^4 + 5x^2 + 4 ✓

4Problem 4medium

❓ Question:

A polynomial of degree 5 has zeros at x = 2, x = -1, and x = 3 + i. What other zeros must it have?

💡 Show Solution

Step 1: Apply Fundamental Theorem of Algebra: A polynomial of degree 5 has exactly 5 zeros (counting multiplicities)

Step 2: List known zeros: • x = 2 • x = -1 • x = 3 + i

Step 3: Apply Complex Conjugate Root Theorem: If a polynomial has real coefficients and 3 + i is a zero, then 3 - i must also be a zero

Step 4: Count zeros so far: 2, -1, 3 + i, 3 - i = 4 zeros

Step 5: Find remaining zero: 5 - 4 = 1 more zero needed This could be any real number or a repeated root

Step 6: Most complete answer: Must have: 3 - i May have: one additional real zero or a repeated root

Answer: Must have 3 - i; needs one more zero (real or repeated)

5Problem 5hard

❓ Question:

Factor completely over the complex numbers: f(x) = x⁴ - 16

💡 Show Solution

Step 1: Recognize difference of squares: x⁴ - 16 = (x²)² - 4² = (x² - 4)(x² + 4)

Step 2: Factor x² - 4 (difference of squares): x² - 4 = (x - 2)(x + 2)

Step 3: Factor x² + 4 (sum of squares): Over complex numbers: x² + 4 = x² - (-4) x² = -4 x = ±2i So: x² + 4 = (x - 2i)(x + 2i)

Step 4: Write complete factorization: f(x) = (x - 2)(x + 2)(x - 2i)(x + 2i)

Step 5: Verify (optional): 4 zeros: 2, -2, 2i, -2i Degree 4 polynomial has 4 zeros ✓

Answer: f(x) = (x - 2)(x + 2)(x - 2i)(x + 2i)

Explain using:

⚠️ Common Mistakes: Fundamental Theorem of Algebra and Factoring

Avoid these 4 frequent errors

🌍 Real-World Applications: Fundamental Theorem of Algebra and Factoring

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values

📌 Related Topics in Polynomial and Rational Functions

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Apply the Fundamental Theorem of Algebra to find all roots of polynomials and completely factor polynomial expressions.
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