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.

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, $-\dfrac{1}{4}$ , −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 = $-\dfrac{1}{4}$ is a root.

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

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 = √-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) = x³ + 2x² − 5x − 6.

This means that x = -1 is a root of x³ + 2x² − 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

Answer

Note: Polynomial equations do not always have "nice" solutions! (By "nice soltuions" 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 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:

3x³ − x² − x + 4 = 0.

Answer

Using LiveMath to find Roots

We can also use LiveMath (or any similar Computer Algebra System) to find roots. This can be done graphically OR using the in-built root-finder.

Using a graph in LiveMath, we can easily find the roots of

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

since the roots are the x-intercepts (i.e. where the function has value 0).

Let's see this how to do this in LiveMath:

LIVEMath

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 LiveMath:

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.

2. Remainder and Factor Theorems

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