# 4. Roots of a Polynomial Equation

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

(a) A polynomial of

n-th degree can be factored intonlinear factors.(b) A polynomial equation of degree

nhas exactlynroots.(c) If `(x − r)` is a

factorof a polynomial, then `x = r` is arootof 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`).

We discussed this example in 3. How to Factor Polynomials, and found the factors to be:

4

x^{3}− 3x^{2}− 25x− 6 = (x− 3)(4x+ 1)(x+ 2)

Recall 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`, `x=-1/4`, and `x= −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:

Graph of *f*(*x*) = 4*x*^{3} − 3*x*^{2} − 25*x* − 6

### Example 2

The equation *x*^{5} − 4*x*^{4} *− *7*x*^{3} + 14*x*^{2} − 44*x* + 120 = 0 can be factored (using Wolfram|Alpha) 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, as expected.

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

Graph of *y* = *x*^{5} − 4*x*^{4} *− *7*x*^{3} + 14*x*^{2} − 44*x* + 120

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

### Example 3

In the earlier section, 2. 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.

Another way to see what's going on is to graph the polynomial.

Graph of *y* = *x*^{3} + 2*x*^{2} − 5*x* − 6

The graph shows us the other 2 roots, −3 and 2.

### Example 4

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

x^{4}+ 0.4x^{3}− 6.49x^{2}+ 7.244x− 2.112 = 0

Answer

Wolfram|Alpha's result is:

Solve: *x*^{4} + 0.4*x*^{3} − 6.49*x*^{2} + 7.244*x* − 2.112 = 0 ,

Solution is: {`x = -3.2`}, {`x = 1.2`}, {`x = 0.5`}, {`x = 1.1`}

Here is the graph:

Graph of *y* = *x*^{4} + 0.4*x*^{3} − 6.49*x*^{2} + 7.244*x* − 2.112

The three positive roots are difficult to see. Here is that portion again, zoomed in for a clearer view:

Graph of *y* = *x*^{4} + 0.4*x*^{3} − 6.49*x*^{2} + 7.244*x* − 2.112

**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 should be seen as an historical approach, 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 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 a computer algebra system:

3

x^{3}−x^{2}−x+ 4 = 0.

Answer

3*x*^{3} − *x*^{2} − *x* + 4, Solution is: {`x = -1.0914`, `x≈0.71237 - 0.84509 i`, `x≈0.71237 + 0.84509 i` }

We see there is one real solution and 2 complex solutions.

Checking this graphically, we have:

Graph of *y* = 3*x*^{3} − *x*^{2} − *x* + 4

We can see that there is only one (real) root, near `x = -1` as expected.

## 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 can 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:

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:

Graph of *y* = *x*^{5} + 8.5*x*^{4} + 10*x*^{3} − 37.5*x*^{2} − 36*x *+ 54.

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.