# 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*) =* *4*x*^{3} − 3*x*^{2} − 25*x* − 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

4

x^{3}− 3x^{2}− 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) =4x^{3}− 3x^{2}− 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.

### Example 2

The equation *x*^{5} − 4*x*^{4} *− *7*x*^{3} + 14*x*^{2} − 44*x* + 120 = 0 can be factored (using Scientific Notebook) and written as:

(

x− 2)(x− 5)(x+ 3)(x^{2}+ 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.

[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*) = *x*^{3} + 2*x*^{2} − 5*x *− 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:

x^{4}+ 0.4x^{3}− 6.49x^{2}+ 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:

3

x^{3}−x^{2}−x+ 4 = 0.

## Using Computers to find Roots

We can also use technology to find roots. This can be done 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:

x^{5}+ 8.5x^{4}+ 10x^{3}− 37.5x^{2}− 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 − 2jandx= 0 + 2j

### Example 7

The factors of the polynomial *x*^{3}+ 7*x*^{2} + 17*x* + 15 are found using a computer algebra system as follows:

x^{3}+ 7x^{2}+ 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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