Example

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

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

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

Relationship Between Roots and Factors

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

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

Multiplicity

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

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: $(x - r)$ where $r$ is a real root
2. Irreducible quadratic factors: $ax^2 + bx + c$ where $b^2 - 4ac < 0$

Linear Factorization Theorem

If $p(x)$ is a polynomial of degree $n \geq 1$, then: $p(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n)$

where $a_n$ is the leading coefficient and $r_1, r_2, \ldots, r_n$ are the roots (possibly complex, possibly repeated).

Finding All Roots

Step-by-step process:

1. Count the roots: Degree $n$ means $n$ 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 $a + bi$, also include $a - bi$

Complex Conjugate Root Theorem

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

Consequence for Factoring

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

This is a real quadratic (no imaginary coefficients).

Writing Polynomials from Roots

Given roots, we can construct the polynomial:

Example: Roots are $2, 3, -1$

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

Polynomial: $p(x) = (x - 2)(x - 3)(x + 1) = x^3 - 4x^2 + x + 6$

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

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

Number of Real vs Complex Roots

For a polynomial of degree $n$ with real coefficients:

• Complex roots come in conjugate pairs
• If $n$ is odd, there is at least one real root
• If $n$ 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 $p(x)$. 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 $p(-x)$.

Example

$p(x) = x^3 - 4x^2 + x + 6$

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

$p(-x) = -x^3 - 4x^2 - x + 6$

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

Step 1: Write factors

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

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

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

Step 2: Combine complex conjugate factors

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

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

$= (x + 1)^2 - i^2 = (x + 1)^2 - (-1) = (x + 1)^2 + 1$

$= x^2 + 2x + 1 + 1 = x^2 + 2x + 2$

Step 3: Form the complete polynomial

$p(x) = (x - 2)^2(x^2 + 2x + 2)$

Step 4: Expand

First: $(x - 2)^2 = x^2 - 4x + 4$

Then multiply: $(x^2 - 4x + 4)(x^2 + 2x + 2)$

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

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

Answer: $p(x) = x^4 - 2x^3 - 2x^2 + 8$

Verification:

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

Step 1: Recognize this as quadratic in form

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

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

So $u = -1$ or $u = -4$

Step 3: Solve for $x$

Case 1: $x^2 = -1$ $x = \pm i$

Case 2: $x^2 = -4$ $x = \pm 2i$

All roots: $x = i, -i, 2i, -2i$

Step 4: Write complete factorization

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

Or in terms of real quadratics:

$(x - i)(x + i) = x^2 + 1$ $(x - 2i)(x + 2i) = x^2 + 4$

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

Answer:

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

Verification:

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