Synthetic division, Descarte’s Rule of Signs, and the Rational Roots Theorem, will all be used to find the roots of a function higher than degree 2. One could use these methods to find the roots in an equation of degree 2 but the best method for those is to simply use the quadratic formula, which was previously explained.

 

Here are the steps on how to find the roots of a function and they will be explained in the next example.

1)      Use Descartes Rule of Signs to find the nature of the roots.

2)      Use the Rational Roots Theorem to find the possible value of the roots.

3)      Plug the results from step 2 into synthetic division and through trial and error get a remainder of zero.

4)      After getting a remainder of zero, use what is leftover from synthetic division to find the remaining roots.

 

Here is a walk through of all the steps.

 

Find all the zero’s of f(x) = x³ -5x² -4x + 20

 

Step 1) Use Descartes Rule of Signs to find the nature of the roots.

Counting the sign changes in f(x), there are 2 or 0 positive roots.

 

Finding f(-x) = -x³ -5x² +4x + 20, realize there is 1 sign change so that means that there is 1 negative root.

 

That means that there is

 

2 positive 1 negative 0 imaginary

or

0 positive 1 negative 2 imaginary

 

 

Step 2) Use the Rational Roots Theorem to find the possible value of the roots.

The possible rational zero’s for this function are ±1,±2,±4,±5,±10,±20.

 

Step 3) Plug the results in from step 2 into synthetic division and through trial and

error until there is a remainder of zero.

 

This means to try the possible real rational zero’s from step 2 and plug them into the upper left hand box in synthetic division to find out which one will get a remainder of 0.

This is all trial and error but it is a good idea to start with the smaller numbers since they are easier to work with. Try the +1 first.

 

 

 

 

 

1	1	-5	-4	20
		1	-4	-8
	1	-4	-8	12

 

 

 

Since 12 is the remainder, 1 is not a root. Now try –1.

 

-1	1	-5	-4	20
		-1	6	-2
	1	-6	2	18

 

 

-1 is not a root. Look at 2.

 

2	1	-5	-4	20
		2	-6	-20
	1	-3	-10	0

The remainder is 0. That means that 2 is a factor.

 

Step 4) After getting a remainder of zero, which means that the number used is a root, use what is leftover from synthetic division to find the remaining roots.

What is leftover is the equation x² -3x – 10 (the bottom row). Set this equal to 0, x² -3x – 10 = 0, and solve using the quadratic formula. This results in –2 and 5 as answers also. Write the roots in order form least to greatest.

-2, 2, 5 are the answers!

 

 

A good understanding of synthetic division is important for the next part. Working backwards, if 2 is a factor then x = 2. Set this equation to 0 by moving the 2 to the left side. x – 2 = 0. Therefore at the end of step 3 in the previous example, one could have rewritten the original polynomial as f(x) = (x-2)(x² – 3x –10) = 0. Factor the last set of parentheses. f(x) = (x-2)(x-5)(x+2). There could be another method to solve it but more important is the understanding that it works like this. If one took the function f(x) = (x-2)(x-5)(x+2) and set each individual parentheses equal to 0, they would have the roots of the equation.

 

If one were to graph the original function, they would find that it crosses the x-axis at precisely the points –2,2, and 5. Following is a graph of that function.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example

 

Find all the roots of f(x) = 12x⁴ +4x³ -3x² – x

 

The first thing to realize is that there is an x in every term. Factor an x out of every term to begin with and instead of setting equal to f(x) go ahead and set it equal to 0.

 

x(12x³ +4x² -3x –1) = 0

 

Since they all had one x in common there is one root at 0. Take the remaining part, 12x³ +4x² -3x –1, and find the roots of it by first determining the nature of the roots.

 

f(x) = 12x³ +4x² -3x –1

 

1 sign change means 1 positive.

 

f(-x) = -12x³ +4x² +3x –1

 

2 sign changes mean 2 or 0 negative roots.

 

Note- The reason to do this step is because there has to be a positive root and there is a possibility of 2 negative roots. If there were 0 positive roots then do not waste time in the next step by plugging in positive numbers. The same can be said for negative roots.

 

Find all the possible rational roots.

 

Plug these numbers into synthetic division to try and get a remainder of 0. Start with 1.

 

1	12	4	-3	-1
		12	16	13
	12	16	13	12

 

 

1 is not a root. Try –1.

 

-1	12	4	-3	-1
		-12	8	5
	12	-8	5	-6

 

 

 

-1 is not a root. Try .

 

1/2	12	4	-3	-1
		6	5	1
	12	10	2	0

 

 

is a root. Leftover is 12x² +10x +2. Solve 12x² +10x +2 = 0.

 

 

and are the results from solving 12x² +10x +2 = 0, using the quadratic formula. List all the roots, remember that 0 is a root because x was factored out at the beginning.

 

 

Answer!

 

 

 

 

Example

Solve x⁵ + x³ +2x² -12x + 8 = 0

 

Use Descarte’s Rule of Signs.

f(x)= x⁵ + x³ +2x² -12x + 8

 

2 sign changes means 2 or 0 positive roots.

 

f(-x)= -x⁵ - x³ +2x² +12x + 8

 

1 sign change means 1 negative root. Find all the possible rational zeros.

±1,±2,±4,±8

 

Plug these into synthetic division to find a root. Remember that the x⁴ term is missing so plug a zero in for it along the top row.

1	1	0	1	2	-12	8
		1	1	2	4	-8
	1	1	2	4	-8	0

 

 

1 is a root. Take what is left, x⁴ + x³ +2x² +4x – 8 and repeat the same step like it is a new problem. There is no reason to redo Descarte’s Rule of Signs and The Rational Zero Theorem since the same results will apply here. The only difference will be the number of positive roots, which are lessened by 1, since 1 is a root. There is another positive root so try 1 again with the new equation.

 

1	1	1	2	4	-8
		1	2	4	8
	1	2	4	8	0

 

 

1 is a root again. That means it is a double root. Left over is x³ +2x² +4x + 8. Try more possible rational zero’s here. It would be a waste of time to try the 1, or any positive number since there are only 2 positive roots and have found out that they are both 1. Try –1.

 

-1	1	2	4	8
		-1	-1	-3
	1	1	3	5

-1 is not a root. Remember all the positive roots have been found, so do not try any more positive numbers. Try –2.

 

-2	1	2	4	8
		-2	0	-8
	1	0	4	0

 

-2 is a root. Now there is a result of degree 2, x² +4. Solve this using the quadratic formula and get ±2i. These are the 2 imaginary roots. List all answers least to greatest with the imaginary numbers at the end.

 

-2,1, ±2i

 

Answer! There is no need to write the 1 twice.

 

Knowing that the roots of x⁵ + x³ +2x² -12x + 8 = 0 are -2,1, and ±2i with 1 as a double root. Rewrite the original equation as

 

(x-1)(x-1)(x+2)(x-2i)(x+2i) = 0.

This understanding is very important in the next problem.

 

 

 

 

Example

Write the equation with roots at –2,1,and 3.

 

Rewrite the problem factored out.

(x+2)(x-1)(x-3) = 0

 

Multiply out the polynomial.

x³ -2x² -5x + 6 = 0

 

Answer!