3.3: Real Zeros of Polynomials

What Are We Doing?

Section 3.1 and Section 3.2 focused on graphs of polynomials.

We now dive into the algebra of polynomials; in particular we will look at how to factor polynomials into their real zeros.

Core Idea

Recall the definition of a zero:

zeros
Suppose $P$ is a polynomial and $c$ is a real number. Then the following are equivalent:

#4 is the key here: if $c$ is a zero of a degree $n$ polynomial $P(x)$, then $P(x)$ admits a factorization \[\underbrace{P(x)}_{\deg n} = \underbrace{(x-c)}_{\deg 1}\cdot \underbrace{Q(x)}_{\deg n - 1}\]

Dividing both sides by $(x-c)$: \[\dfrac{P(x)}{x - c} = Q(x)\]

So in order to factor, we need to learn how to divide polynomials.

Long Division of Polynomials

Dividing polynomials is the same as dividing numbers.

Consider the number $37$. If we divide by 7, we can write \[37 = 7\cdot 5 + 2\] or dividing both sides by $7$: \[\dfrac{37}{7} = 5 + \dfrac{2}{7}\]

Polynomial division is the same as the previous example, except the numbers are replaced with polynomials.

division algorithm
$P(x)$ and $D(x)$ are polynomials, with $D(x) \neq 0$. Then there exist unique polynomials $Q(x)$ and $R(x)$, where the degree of $R(x)$ is 0 or less than the degree of $P(x)$ where: \[\dfrac{P(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)} \qquad \text{ or } \qquad P(x) = D(x)\cdot Q(x) + R(x)\]

The second form is more useful, we will call it "one-line" notation.

Divide $6x^2 - 26x + 12$ by $x - 4$ and put the result in one-line notation.
Divide $8x^4 + 6x^2 - 3x + 1$ by $2x^2 - x + 2$ and put the result in one-line notation.

Remainder and Factor Theorems

Now we will learn how to factor by dividing.

Remainder Theorem
If $P(x)$ is divided by $x - c$, then the remainder is $P(c)$.
If $P(x) = 3x^5 + 5x^4 - 4x^3 + 7x + 3$ is divided by $(x + 1)$, what is the remainder?

Remember that if $c$ is a zero of $P(x)$, then $P(c) = 0$.

We just saw $P(c)$ is the remainder, and in particular, we just wrote down it has to be zero! Therefore:

Factor Theorem
$c$ is a zero of $P$ if and only if $(x-c)$ is a factor of $P(x)$.
Suppose $P(x) = x^3 - 7x + 6$. Show that $P(1) = 0$, and use this fact to completely factor $P(x)$.

There are two types of "complete factorization."

The first one is over $\mathbb{R}$:

Complete Factorization of $P(x)$ over $\mathbb{R}$
A complete factorization of a polynomial $P(x)$ over $\mathbb{R}$ is one where the resultant factors only have real coefficients.

Here's how to find a complete factorization over $\mathbb{R}$.

Complete Factorization of $P(x)$ over $\mathbb{R}$
For each zero $c$:
  1. Setup:Convert into a factor $(x - c)$.
  2. Divide: Use the division algorithm to divide $P(x) = (x - c)Q(x)$.
  3. Multiplicity: Check if $c$ is a zero of $Q(x)$. If so, repeat and keep dividing $Q(x)$ until $c$ is no longer a zero. Move on to the next zero.
Repeat until you have a quadratic. Then try to factor using Chapter 1 techniques ("new" X method, or the quadratic formula).
Do not factor irreducibles, leave them alone.
Suppose $P(x) = x^2 - x - 1$. Find a complete factorization over $\mathbb{R}$.

If $c$ is a zero, by definition $x = c$ is a solution of $x^2 - x - 1 = 0$.

Using the quadratic formula with $a = 1, b = -1, c = -1$: \begin{align} x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\&= \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4\cdot 1 \cdot (-1)}}{2\cdot 1} \\&= \dfrac{1 \pm \sqrt{1 + 4}}{2} \\&= \dfrac{1 \pm \sqrt{5}}{2} \end{align}

The solutions are $x = \dfrac{1 + \sqrt{5}}{2}$ and $x = \dfrac{1 - \sqrt{5}}{2}$.

Therefore $x^2 -x - 1 = \boxed{\left(x - \dfrac{1 + \sqrt{5}}{2}\right)\left(x - \dfrac{1 - \sqrt{5}}{2}\right)}$

Suppose $P(x) = x^4 - x^3 - 2x^2 + x + 1$. Verify $P(1) = 0$ and $P(-1) = 0$ and find a complete factorization over $\mathbb{R}$.

Structure of Complete Factorization over $\mathbb{R}$

A complete factorization over $\mathbb{R}$ could include irreducible quadratics:

irreducible
An irreducible polynomial is a quadratic polynomial with no real zeros.
An irreducible's discriminant $b^2 - 4ac < 0$ due to the quadratic formula \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] The $\sqrt{b^2 - 4ac}$ is a root of a negative number, so the zeros can't be real!
Determine whether $P(x) = x^2 + 1$ is irreducible.

Because irreducibles don't have real zeros, we don't want them in a complete factorization over $\mathbb{R}$.

Therefore, we leave irreducibles alone when finding a complete factorization over $\mathbb{R}$.

This tells us the structure of a complete factorization over $\mathbb{R}$:

Real Factorization Theorem

Suppose $P(x)$ is a polynomial with real coefficients. A complete factorization of $P(x)$ over $\mathbb{R}$ will break down into linear (degree 1) factors and irreducible quadratics.

There are three possibilities: \begin{align} P(x) &= (\text{linear factors}) \\P(x) &= (\text{linear factors}) \cdot (\text{irreducible factors}) \\P(x) &= (\text{irreducible factors}) \end{align}
Is $(x-3)(2x-5)(x^2 - 4x - 2)$ a complete factorization over $\mathbb{R}$?
Suppose $P(x) = x^4 - 4x^3 + 5x^2 - 4x + 4$. Verify $P(2) = 0$ and find a complete factorization over $\mathbb{R}$.

First, we verify $P(2) = 0$. We have \begin{align}P(2) &= 2^4 - 4\cdot 2^3 + 5\cdot 2^2 - 4\cdot 2 + 4 \\&= 16 - 32 + 20 - 8 + 4 \\&= -16 + 12 + 4 \\&= 0\end{align}

So $(x - 2)$ is a factor. Long dividing: So $Q(x) = x^3 - 2x^2 + x - 2$.

We have two options:

  1. Multiplicity: check if $x = 2$ is a factor of $x^3 - 2x^2 + x - 2$.
  2. Chapter 1: we could just do grouping on $Q(x)$.

Let's go with option 2. We have \begin{align} Q(x) &= x^3 - 2x^2 + x - 2 \\&= x^2(x - 2) + (x -2) \\&= (x-2)(x^2 + 1) \end{align}

Because we saw earlier $x^2 + 1$ is irreducible, a complete factorization of $P(x)$ over $\mathbb{R}$ is $\boxed{P(x) = (x-2)^2(x^2+1)}$

Section 3.5 will show you the irreducibles can be factored!