# 3. Factors and Roots of a Polynomial Equation

Here are three important theorems relating to the roots of a polynomial:

(a) A polynomial of n-th degree can be factored into n linear factors.

(b) A polynomial equation of degree n has exactly n roots.

(c) If (x − r) is a factor of a polynomial, then x = r is a root of the associated polynomial equation.

Let's look at some examples to see what this means.

### Example 1

The cubic polynomial f(x) = 4x3 − 3x2 − 25x − 6 has degree 3 (since the highest power of x that appears is 3).

This polynomial can be factored (using Scientific Notebook or similar software) and written as

4x3 − 3x2 − 25x − 6 = (x − 3)(4x + 1)(x + 2)

So we see that a 3rd degree polynomial has 3 roots.

The associated polynomial equation is formed by setting the polynomial equal to zero:

f(x) = 4x3 − 3x2 − 25x − 6 = 0

In factored form, this is:

(x − 3)(4x + 1)(x + 2) = 0

We see from the expressions in brackets and using the 3rd theorem from above, that there are 3 roots, x = 3, -1/4,  −2.

In this example, all 3 roots of our polynomial equation of degree 3 are real.

Since (x − 3) is a factor, then x = 3 is a root.

Since (4x + 1) is a factor, then x=-1/4 is a root.

Since (x + 2) is a factor, then x = −2 is a root.

Here's the graph of our polynomial, showing the x-intercepts, which are the roots:

### Example 2

The equation x5 − 4x4 7x3 + 14x2 − 44x + 120 = 0 can be factored (using Scientific Notebook) and written as:

(x − 2)(x − 5)(x + 3)(x2 + 4) = 0

We see there are 3 real roots x = 2, 5, -3, and 2 complex roots x = ±2j, (where j =sqrt(-1)).

So our 5th degree equation has 5 roots altogether.

On the graph, we can see the three real roots only:

[Do you need revision on complex numbers? Go to Complex Numbers.]

### Example 3

In the previous section, Remainder Theorem and the Factor Theorem, we found in one of the examples that (x + 1) is a factor of f(x) = x3 + 2x2 − 5x − 6.

This means that x = -1 is a root of x^3+ 2x^2− 5x − 6 = 0.

[To check this, substitute x = -1 into the polynomial. If it is a root, then you should get value 0 when you substitute.]

### Example 4

The following polynomial equation would be rather tricky to solve using the Remainder and Factor Theorems. We will solve it using Scientific Notebook:

x4 + 0.4x3 − 6.49x2 + 7.244x − 2.112 = 0

Note: Polynomial equations do not always have "nice" solutions! (By "nice solutions" I mean solutions which are integers or simple fractions.) This is why I feel the Remainder and Factor Theorems are pretty useless, because you can only use them if at least some of the solutions are integers or simple fractions.

If you use a computer algebra system (like Wolfram | Alpha or Scientific Notebook) to solve these, you can be done in seconds and move on to something more meaningful, like the applications.

### Example 5

Solve the following polynomial equation using Scientific Notebook:

3x3x2x + 4 = 0.

## Using a Computer Algebra System to find Roots

We've been using technology to find most of the roots above. This is better than trying guess solutions and then dividing polynomials. Using a computer, we cna quickly find the roots either graphically OR using the in-built root-finder when available.

Using a graph, we can easily find the roots of polynomial equations that don't have "nice" roots, like the following:

x5 + 8.5x4 + 10x3 − 37.5x2 − 36x + 54 = 0.

The roots of the equation are simply the x-intercepts (i.e. where the function has value 0). Here's the graph of the function:

We can see the solutions are x=-6, x=-3, x=-2, x=1 and x=1.5. (Zooming in close to these roots on the graph confirms these values.)

## Complex Roots

Regarding complex roots, the following theorem applies :

If the coefficients of the equation f(x)=0 are real and a + bj is a complex root, then its conjugate a − bj is also a root.

For more on complex numbers, see: Complex Numbers

### Example 6

In Example (2) above, we had 3 real roots and 2 complex roots. Those complex roots form a complex conjugate pair,

x = 0 − 2j and x = 0 + 2j

### Example 7

The factors of the polynomial x3+ 7x2 + 17x + 15 are found using a computer algebra system as follows:

x3 + 7x2 + 17x + 15 = (x + 3)(x + 2 − j)(x + 2 + j)

So the roots are

x = −3

x = −2 + j and

x = −2 − j

There is one real root and the remaining 2 roots form a complex conjugate pair.

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