Fundamental Theorem of Algebra and Factoring

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

Fundamental Theorem of Algebra and Factoring

The Fundamental Theorem of Algebra

Theorem: Every polynomial equation of degree n1n \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)=x36x2+11x6p(x) = x^3 - 6x^2 + 11x - 6 is degree 3, so it has exactly 3 roots.

If we factor: p(x)=(x1)(x2)(x3)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 (xr)(x - r) is a factor of p(x)p(x).

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

Multiplicity

If (xr)k(x - r)^k is a factor but (xr)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: (xr)(x - r) where rr is a real root
  2. Irreducible quadratic factors: ax2+bx+cax^2 + bx + c where b24ac<0b^2 - 4ac < 0

Linear Factorization Theorem

If p(x)p(x) is a polynomial of degree n1n \geq 1, then: p(x)=an(xr1)(xr2)(xrn)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 abia - bi

Complex Conjugate Root Theorem

Theorem: If a polynomial has real coefficients and a+bia + bi (where b0b \neq 0) is a root, then the complex conjugate abia - bi is also a root.

Consequence for Factoring

If a+bia + bi and abia - bi are roots, the corresponding quadratic factor is: [x(a+bi)][x(abi)]=x22ax+(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: (x2)(x3)(x+1)(x - 2)(x - 3)(x + 1)

Polynomial: p(x)=(x2)(x3)(x+1)=x34x2+x+6p(x) = (x - 2)(x - 3)(x + 1) = x^3 - 4x^2 + x + 6

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

p(x)=(x1)[x(2+i)][x(2i)]p(x) = (x - 1)[x - (2 + i)][x - (2 - i)] =(x1)[x24x+5]= (x - 1)[x^2 - 4x + 5] =x35x2+9x5= 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)=x34x2+x+6p(x) = x^3 - 4x^2 + x + 6

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

p(x)=x34x2x+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)=x37x2+14x8p(x) = x^3 - 7x^2 + 14x - 8 and write the complete factorization.

💡 Show Solution

Solution:

Given: p(x)=x37x2+14x8p(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)=17+148=0p(1) = 1 - 7 + 14 - 8 = 0

So x=1x = 1 is a root, and (x1)(x - 1) is a factor.

Step 2: Divide by (x1)(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=1ix = -1 - i (complex conjugate pair)

Step 1: Write factors

From x=2x = 2 (multiplicity 2): (x2)2(x - 2)^2

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

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

Step 2: Combine complex conjugate factors

[x(1+i)][x(1i)]=[(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: (ab)(a+b)=a2b2(a - b)(a + b) = a^2 - b^2

=(x+1)2i2=(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)=(x2)2(x2+2x+2)p(x) = (x - 2)^2(x^2 + 2x + 2)

Step 4: Expand

First: (x2)2=x24x+4(x - 2)^2 = x^2 - 4x + 4

Then multiply: (x24x+4)(x2+2x+2)(x^2 - 4x + 4)(x^2 + 2x + 2)

=x4+2x3+2x24x38x28x+4x2+8x+8= x^4 + 2x^3 + 2x^2 - 4x^3 - 8x^2 - 8x + 4x^2 + 8x + 8

=x42x32x2+8= x^4 - 2x^3 - 2x^2 + 8

Answer: p(x)=x42x32x2+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)=(xi)(x+i)(x2i)(x+2i)p(x) = (x - i)(x + i)(x - 2i)(x + 2i)

Or in terms of real quadratics:

(xi)(x+i)=x2+1(x - i)(x + i) = x^2 + 1 (x2i)(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)=(xi)(x+i)(x2i)(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