To solve equations of degree 3 and higher, it is easiest to use synthetic division. Although tricky at first, after working a few problems the process is relatively quite simple. Before explaining the process, here is an overview.

For example, 3x⁴ – 6x³ –2x² + x – 6 = 0. In solving this equation look for all the different values of x that make this equation true. In other words, try to find out what numbers when substituted for x in the equation, will make it true. Since the highest exponent is 4 in this equation, one would say it is of degree 4 and therefore there are 4 values that will work for x.

 

Note- The degree will always state the number of solutions.

 

Before solving this type of equation, it is important to learn how to use synthetic division. Synthetic division is used to divide one polynomial by another polynomial.

Example

 

Divide 3x⁴ – 6x³ –2x² + x – 6 by x + 1.

 

The first thing to do here is to take, x+1, set it equal to 0, and solve for x.

x+1 = 0

x = -1

 

Now construct the following diagram

 

 

 

 

 

 

 

In the upper left hand corner, put the result from the previous step, x = -1.

 

1

 

 

Take all the coefficients, the numbers in front of the x’s, and put them in a straight row left to right across the top. Make sure to put these in order. If there is a degree that is missing, put a zero in it’s place, this will be covered more later on.

 

-1		3	-6	-2	1	-6

 

 

 

Make sure when setting this up, the last coefficient, -6 is at the end, directly over the box on bottom.

Take the first coefficient, 3, and bring it down.

 

-1		3	-6	-2	1	-6
		3

 

 

Next is to take the –1 in the box and multiply it by the 3, this results in –3. Take that number and put it directly underneath the next coefficient in the row, -6.

 

 

-1		3	-6	-2	1	-6
			-3
		3

 

 

Take the –6 and add whatever is directly underneath it, in this case it is the -3. Write the result underneath the -3 in that column.

-1		3	-6	-2	1	-6
			-3	9
		3	-9

 

 

 

Do not get intimidated and confused, if having trouble with the preceding steps read back over them make sure they are understood. From here, continue with the same process. Take the –9 and multiply it by the number in the box, -1, resulting in 9. Place the 9 directly underneath the next coefficient, -2.

-1 3 -6 -2 1 -6

3 9

 

3 -9

 

 

Add the –2 and 9 and write the result underneath the 9.

 

-1		3	-6	-2	1	-6
			-3	9
		3	-9	7

 

Add 1 and –7, write the result underneath the –7.

 

-1		3	-6	-2	1	-6
			-3	9	-7
		3	-9	7	-6

 

Multiply the –6 by the –1 in the box, write the result, 6, underneath the next coefficient.

 

-1		3	-6	-2	1	-6
			-3	9	-7	6
		3	-9	7	-6

 

Add –6 and 6, put the result in the last box on bottom.

 

-1		3	-6	-2	1	-6
			-3	9	-7	6
		3	-9	7	-6	0

 

Look at the bottom row in the last step of synthetic division, 3 -9 7 -6 and in the box, 0. The answer can be obtained by using these numbers. First, ignore the number in the right hand box, the 0, it is the remainder. Remember that this problem started out with a polynomial of degree 4, subtract 1 from that number and get three, this is the degree of the answer. Now rewrite the bottom numbers in order as a polynomial of degree 3.

3x³ –9x² + 7x – 6 is the answer.

 

Try to understand what each number in the previous problem means and it will greatly increase the understanding of this concept and how to use it later. First, think about the simplified problem 12 divided by 4. The answer is 3. That means that 3(4) = 12. Now apply that to this problem, if 3x⁴ – 6x³ –2x² + x – 6 divided by x + 1 is 3x³ –9x² + 7x – 6 then it only goes to reason that

(3x³ –9x² + 7x – 6)(x + 1) = 3x⁴ – 6x³ –2x² + x – 6.

 

Note- In this problem the remainder was 0. The number underneath the right hand box is always the remainder. When the remainder is not 0 then there are a few adjustments on writing the answer that one has to do, but that will be covered in later problems.

 

 

 

 

Example

Divide x² + 3x + 5 by x + 1

 

Solve x + 1 = 0 and get x = -1.

 

-1		1	3	5

 

Construct the diagram. Bring down the 1.

 

-1		1	3	5
		1

 

Multiply 1 and –1. Put the result underneath the 3.

 

-1		1	3	5
			-1
		1

 

Add the 3 and –1. Put the result underneath the –1.

 

 

-1		1	3	5
			-1
		1	2

Multiply the 2 and –1. Put the result underneath the 5.

-1		1	3	5
			-1	-2
		1	2

 

Add the 5 and –2. The result goes in the box.

 

-1		1	3	5
			-1	-2
		1	2	3

 

Write the answer in proper form. Notice that there is a remainder in this case, it is 3.

 

Take the bottom row of synthetic division and drop one degree, in this case from 2 to 1, and rewrite the polynomial. What makes this one a little different is the fact that the remainder is not 0 here it is 3. All one has to do with the remainder is to rewrite it over what the problem was originally dividing by, x+1.

 

x + 2 +

 

 

 

Answer!

 

 

 

Knowing this, state that x² + 3x + 5 = (x + 2 + ) (x + 1).

 

 

 

 

 

Example

 

Divide 2x⁴ – 5x³ + 2x – 3 by x –1

 

Note- The first thing to notice in this problem is not only that it is of degree 4 but also the x² term is missing. Account for that term whether it is there or not, so when constructing the diagram make sure and put a 0 in the place of that term.

 

Solve x-1 = 0, and get x =1.

 

1	2	-5	0	2	-3

 

Construct the diagram and bring down the 2.

1	2	-5	0	2	-3
	2

 

Multiply the 1 and 2.

 

1	2	-5	0	2	-3
		2
	2

 

Add –5 and 2. Multiply the result by 1.

 

1	2	-5	0	2	-3
		2	-3
	2	-3

 

Add 0 and –3 and multiply the result by 1.

 

1	2	-5	0	2	-3
		2	-3	-3
	2	-3	-3

 

Add 2 and –3. Multiply the result by 1.

 

1	2	-5	0	2	-3
		2	-3	-3	-1
	2	-3	-3	-1	-4

 

Rewrite the answer.

 

2x³ –3x² -3x – 1 -

 

Answer!