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.
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:
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.]
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.]
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.
Solve the following polynomial equation using Scientific Notebook:
3x3 − x2 − x + 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.)
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
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
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.