In past courses, you have had a good deal of experience working with polynomial equations of degree 1 (linear) and degree 2 (quadratic). Unfortunately, when the degree of a polynomial equation increases, the equation becomes more difficult to solve. Finding the roots of an equation such as x5 + x4 - 5x3 - x2 + 8x - 4 = 0 can be a difficult task. In this course, we will just be touching the surface on techniques for solving higher degree polynomial equations. Our emphasis will be on using what we already know to help us solve new situations.

FYI: The fifth degree polynomial equation shown above factors into (x - 1)3 (x + 2)2 = 0.

A question requiring that you solve a polynomial equation may be worded in a variety of formats.
Consider the following:
• Solve the polynomial equation P(x) = 0.
• Find the real or imaginary solutions of the polynomial equation P(x) = 0.
• Find the roots of the polynomial equation P(x) = 0.
• Find the zeros of the polynomial function P(x).
• Factor the polynomial function P(x) = 0 and express the roots.

These statements are different ways of
asking the same thing!
 confused

So, how many solutions (roots) does a polynomial equation possess?
According to the Fundamental Theorem of Algebra, a polynomial of degree n will have n roots, some of which may be multiple roots (repeated roots) or complex roots (conjugate pairs).
For example, x5 + x4 - 5x3 - x2 + 8x - 4 = 0 is a polynomial of degree 5 (highest power) and as such will have 5 roots. This equation factors to (x - 1)3 (x + 2)2 = 0.
There are actually five factors: (x - 1)(x - 1)(x - 1)(x + 2)(x + 2) = 0.
The factor of (x - 1) has a multiplicity of 3.
The factor of (x + 2) has a multiplicity of 2.
While there are five roots, x = 1, x = 1, x = 1, x = -2, and x = -2, we can see
that the root x = 1 is repeated three times and x = -2 is repeated twice.
It is customary to say that the solution is simply 1 and -2.

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Let's take a look at some strategies for solving higher degree polynomial equations.

expin1 Strategy: Find the common factor; use the Quadratic Formula.
Find the solutions to the equation 4x4 + 16x2 = 8x3.
4x4 - 8x3 + 16x2 = 0
Rearrange to P(x) = 0 form.
4x2(x2 - 2x + 4) = 0
Factor out greatest common factor of 4x2.
4x2 = 0;     x2 - 2x + 4 = 0
Set factors equal to zero (Zero Product).
4x2 = 0 implies 4 • x • x = 0
4 ≠ 0;  x = 0;  x = 0 (repeated)

x2 - 2x + 4 = 0 needs the Quadratic Formula to solve.

polyquadf
The 4 Solutions:   polysolution4 (the repeated root need not be listed twice)

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expin2 Strategy: Look for a quadratic-style pattern; employ substitution.
Find the real or imaginary solutions to x4 = 13x2 - 36.
x4 - 13x2 + 36= 0
Rearrange to P(x) = 0 form.
polypattern
Notice a pattern: the variable x2 being squared and being used to a power of one. Get in the habit of looking for this quadratic-style pattern.
Let m = x2
Set x2 = m (or any variable of your choosing).
This process may help you to see the rest of the solution to this problem. Make the substitutions.
polypat2
polypat3
You now have a quadratic equation that can be easily solved by factoring.
Be careful not to stop when you solve for m. Remember that m is really x2.
x2 = 9;     x2 = 4
Replace m with x2 and solve for the answers to the original equation.
The 4 Solutions:    x = ±3;      x = ± 2

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expin3 Strategy: Common factor; quadratic-style pattern; substitution.
Find the zeros of the polynomial function P(x) = x5 - 12x3 + 32x.
x5 - 12x3 + 32x = 0
Set P(x) = 0.
x(x4 - 12x2 + 32) = 0
Factor out the common factor of x.
polyex3
Notice the pattern: This is the same type of quadratic-style pattern we saw in example 2.
We can see that one solution will be x = 0. So, let's concentrate on the portion inside the parentheses to find the remaining solutions.
Let m = x2
polyex3a
Now, we can solve the quadratic equation by factoring.
Using these factors, we can rewrite the original equation as:
polyex3b
x = 0
x2 - 8 = 0
x2 - 4 = 0
Set the factors equal to zero and solve for x.
The 5 Solutions:    polyex3ans

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expin4 Strategy: Use factoring by grouping.
Solve: a3 + 5a2 - 6a - 30 = 0
(a3 + 5a2) - (6a + 30) = 0
There is no GCF, so group the terms. Be careful of the negative sign in front of the second grouping.
a2(a + 5) - 6(a + 5) = 0
Factor the groupings.
(a + 5)(a2 - 6) = 0
Factor out the common parentheses.
(a + 5) = 0; (a2 - 6) = 0
Set the factors equal to zero.
6ans2
Solve for a.
6ans
The 3 Solutions:   polyeq4red

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expin5 Strategy: Factor by grouping; use difference of perfect cubes.
Solve: x4 + 2x3 - 8x - 16 = 0
(x4 + 2x3) - (8x + 16) = 0
There is no GCF, so group the terms. Be careful of the negative sign in front of the second grouping.
x3(x + 2) - 8(x + 2) = 0
Factor the groupings.
(x + 2)(x3 - 8) = 0
Factor out the common parentheses.
(x + 2)(x - 2)(x2 + 2x + 4) = 0
Be careful! You are not finished. The (x3 - 8) is the difference of perfect cubes and can be factored further.
x + 2 = 0; x - 2 = 0;
x2 + 2x + 4 = 0
Set the factors equal to zero and solve for x.
When solving for x, use the quadratic formula for x2 + 2x + 4 = 0.
a = 1; b = 2; c = 4
quadans
Note: If you had stopped at
(x + 2)(x3 - 8) = 0 and set each factor = 0, getting x3 = 8, and x = 2, you would not have found the complex roots of this equation.
The 4 Solutions:   -2;   2;    irad3

hintgal
Remember!
a3 + b3 = (a + b)(a2 - ab + b2) (factoring sum of cubes)
a3 - b3 = (a - b)(a2 + ab + b2) (factoring difference of cubes)

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