Synthetic Division with Higher Powers

Most high school algebra textbooks cover two methods for dividing polynomials: long division and synthetic division.  The common teaching pattern in these textbooks is to show long division as the default method that works for all cases, and then to show synthetic division as a handy shortcut that only works when the divisor is a first degree polynomial in the form of x – c. 

However, synthetic division can be adjusted to be used for divisors of higher powers, which for some reason is not taught in high school algebra textbooks.  To show how this can be done, we must first examine how synthetic division of first degree divisors relates to long division, and use the same logic to apply the method to higher degree divisors.

First Degree Divisor

Let’s say we were divide x – 2 into 3x⁴ – 4x³ – 8x² + 11x + 1.  The long division method would be as follows:

	3x3	+ 2x2	– 4x	+ 3
x – 2	3x4	– 4x3	– 8x2	+ 11x	+ 1
	3x4	– 6x3
		2x3	– 8x2
		2x3	– 4x2
			– 4x2	+ 11x
			– 4x2	+ 8x
				3x	+ 1
				3x	– 6
					7

If we were to re-write the problem without the repetitive parts (the first terms subtracting to zero, the dropped down terms, and the x’s) we would have the following:

	3	2	–4	3
– 2	3	–4	–8	11	1
		–6
		2
			–4
			–4
				8
				3
					–6
					7

We can further re-write the problem in three lines to save space:

	3	2	–4	3
– 2	3	–4	–8	11	1
		–6	–4	8	–6
		2	–4	3	7

Once again we can eliminate repetition by combining the first and last rows:

– 2	3	–4	–8	11	1
		–6	–4	8	–6
	3	2	–4	3	7

Finally, we can multiply the top left number and every second row number by -1, giving us the following:

2	3	–4	–8	11	1
		6	4	–8	6
	3	2	–4	3	7

We should now recognize that each second row number can be obtained by multiplying the number in the top left by the number in the previous column’s last row, for example, 2 x 3 = 6, 2 x 2 = 4, 2 x -4 = -8, and 2 x 3 = 6.  Also, each number in the last row can be obtained by adding the first and second rows together, for example, 3 + 0 = 3, -4 + 6 = 2, -8 + 4 = -4, 11 + -8 = 3, and 1 + 6 = 7.  We have now transformed our long division problem into a synthetic division problem, a method in which you alternate between adding columns and multiplying by the c value in the divisor x – c.

For example, if we were to use synthetic division to divide x – 2 into x⁴ – x³ – 19x² + 49x – 30, we would start by writing the divisor number opposite 2 in the top left corner and the dividend coefficients 1, -1, -19, 49, and -30 along the top:

2	1	–1	–19	49	-30

Bring down the 1 in the first column:

2	1	–1	–19	49	-30
	1

Multiply 2 x 1 = 2 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2
	1

Add -1 + 2 = 1 in the second column:

2	1	–1	–19	49	-30
		2
	1	1

Multiply 2 x 1 = 2 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2	2
	1	1

Add -19 + 2 = -17 in the third column:

2	1	–1	–19	49	-30
		2	2	–34	30
	1	1	–17	15	0

Multiply 2 x -17 = -34 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2	2	–34
	1	1	–17

Add 49 + -34 = 15 in the fourth column:

2	1	–1	–19	49	-30
		2	2	–34
	1	1	–17	15

Multiply 2 x 15 = 30 and place this number in the second row of the next column:

2	1	–1	–19	49	-30
		2	2	–34	30
	1	1	–17	15

Add -30 + 30 = 0 in the last column:

2	1	–1	–19	49	-30
		2	2	–34	30
	1	1	–17	15	0

The numbers in the last row are the coefficients of the quotient with a degree that is one less than the dividend, which in this case is 1x³ + 1x² – 17x + 15 + ⁰/_{(x – 2)} or x³ + x² – 17x + 15.

Second Degree Divisor

Now we will use the same logic to transform a long division problem into a synthetic division problem when the divisor has a higher degree, for example,  x² – 2x + 3 into 3x⁴ – 4x³ – 8x² + 11x + 1.  The long division method would be as follows:

	3x2	+ 2x	– 13
x2 – 2x + 3	3x4	– 4x3	– 8x2	+ 11x	+ 1
	3x4	– 6x3	+ 9x2
		2x3	– 17x2	+ 11x
		2x3	– 4x2	+ 6x
			– 13x2	+ 5x	+ 1
			– 13x2	+ 26x	– 39
				– 21x	+ 40

If we were to re-write the problem without the repetitive parts (the first terms subtracting to zero, the dropped down terms, the intermediate subtraction steps, and the x’s) we would have the following:

	3	2	– 13
– 2,  3	3	– 4	– 8	11	1
		– 6	9
		2
			– 4	6
			– 13
				26	– 39
				– 21	40

We can further re-write the problem in four lines to save space:

	3	2	– 13
– 2,  3	3	– 4	– 8	11	1
			9	6	– 39
		– 6	– 4	26
		2	– 13	– 21	40

Once again we can eliminate repetition by combining the first and last rows:

– 2,  3	3	– 4	– 8	11	1
			9	6	– 39
		– 6	– 4	26
	3	2	– 13	– 21	40

Finally, we can multiply the top left number and every second and third row number by -1, giving us the following:

2,  – 3	3	– 4	– 8	11	1
			– 9	– 6	39
		6	4	– 26
	3	2	– 13	– 21	40

We should now recognize that each second and third row numbers can be obtained by multiplying the two numbers in the top left by the number in the previous column’s last row, for example, (2, -3) x 3 = (6, -9), (2, -3) x 2 = (4, -6), and (2, -3) x -13 = (-26, 39).  Also, each number in the last row can be obtained by adding the first, second, and third rows together, for example, 3 + 0 + 0 = 3, -4 + + 0 + 6 = 2, -8 + -9 + 4 = -13, 11 + -6 + -26 = -21, and 1 + 39 + 0 = 40.  We have now transformed our long division problem into a synthetic division problem, a method in which you alternate between adding columns and multiplying by the (b, c) value in the divisor x² – bx – c.

For example, if we were to use synthetic division to divide x² – 3x + 2 into x⁴ – x³ – 19x² + 49x – 30, we would start by writing the divisor number opposites 3, -2 in the top left corner and the dividend coefficients 1, -1, -19, 49, and -30 along the top:

3,  – 2	1	– 1	– 19	49	– 30

Bring down the 1 in the first column:

3,  – 2	1	– 1	– 19	49	– 30
	1

Multiply (3, -2) x 1 = (3, -2) and place these two numbers diagonally in the second and third rows of the next two columns:

3,  – 2	1	– 1	– 19	49	– 30
			– 2
		3
	1

Add -1 + 3 = 2 in the second column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2
		3
	1	2

Multiply (3, -2) x 2 = (6, -4) and place these two numbers diagonally in the second and third rows of the next two columns:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4
		3	6
	1	2

Add -19 + -2 + 6 = -15 in the third column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4
		3	6
	1	2	– 15

Multiply (3, -2) x -15 = (-45, 30) and place these two numbers diagonally in the second and third rows of the next two columns:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4	30
		3	6	– 45
	1	2	– 15

Add 49 + -4 + -45 = 0 in the fourth column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4	30
		3	6	– 45
	1	2	– 15	0

Finally, add -30 + 30 = 0 in the last column:

3,  – 2	1	– 1	– 19	49	– 30
			– 2	– 4	30
		3	6	– 45
	1	2	– 15	0	0

The numbers in the last row are the coefficients of the quotient with a degree that is two less than the dividend, which in this case is 1x² + 2x – 15 + ^{(0x + 0)}/_{(x2 – 3x + 2)} or x² + 2x – 15.

Conclusion

Therefore, despite the caution of many algebra textbooks, synthetic division can be adjusted and used for divisors of higher powers.  This article showed synthetic division only for divisors of degree 1 and 2, but the same pattern can be extended to divisors of higher degrees.