SYNTHETIC DIVISION: METHOD AND SOLVED EXERCISES - MATHS - 2025

The synthetic division is a simple way of dividing a polynomial P (x) any one of the form d (x) = x - c. For example, the polynomial P (x) = (x ⁵ + 3x ⁴ -7x ³ + 2x ² -8x + 1) can be represented as the multiplication of the two simplest polynomials (x + 1) and (x ⁴ + 2x ³).

It is a very useful tool since, in addition to allowing us to divide polynomials, it also allows us to evaluate a polynomial P (x) at any number c, which in turn tells us precisely if said number is a zero of the polynomial or not.

Thanks to the division algorithm, we know that if we have two non-constant polynomials P (x) and d (x), there are unique polynomials q (x) and r (x) such that it is true that P (x) = q (x) d (x) + r (x), where r (x) is zero or less than q (x). These polynomials are known as quotient and remainder or remainder respectively.

On the occasions when the polynomial d (x) is of the form x- c, synthetic division gives us a short way of finding who q (x) and r (x) are.

Synthetic division method

Let P (x) = a _n x ⁿ + a _n-1 x ^n-1 +… + a ₁ x + a ₀ the polynomial we want to divide and d (x) = xc the divisor. To divide by the synthetic division method we proceed as follows:

1- We write the coefficients of P (x) in the first row. If any power of X does not appear, we put zero as its coefficient.

2- In the second row, to the left of a _n we place c, and we draw division lines as shown in the following figure:

3- We lower the leading coefficient to the third row.

In this expression b _n-1 = a _n

4- We multiply c by the leading coefficient b _n-1 and we write the result in the second row, but one column to the right.

5- We add the column where we write the previous result and we place the result below that sum; that is, in the same column, third row.

When adding, we have as a result _n-1 + c * b _n-1, which for convenience we will call b _n-2

6- We multiply c by the previous result and write the result to its right in the second row.

7- We repeat steps 5 and 6 until we reach the coefficient at ₀.

8- We write the answer; that is, the quotient and the remainder. Since we are dividing a polynomial of degree n by a polynomial of degree 1, we have that the quotient would be of degree n-1.

The coefficients of the quotient polynomial will be the numbers in the third row except the last one, which will be the residual polynomial or remainder of the division.

Solved exercises

- Example 1

Perform the following division by the synthetic division method:

(x ⁵ + 3x ⁴ -7x ³ + 2x ² -8x + 1): (x + 1).

Solution

We first write the coefficients of the dividend as follows:

Then we write c on the left side, in the second row, along with the dividing lines. In this example c = -1.

We lower the leading coefficient (in this case b _n-1 = 1) and multiply it by -1:

We write its result to the right in the second row, as shown below:

We add the numbers in the second column:

We multiply 2 by -1 and write the result in the third column, second row:

We add in the third column:

We proceed in the same way until we reach the last column:

Thus, we have that the last number obtained is the remainder of the division, and the remaining numbers are the coefficients of the quotient polynomial. This is written as follows:

If we want to verify that the result is correct, it is enough to verify that the following equation is true:

P (x) = q (x) * d (x) + r (x)

So we can check that the result obtained is correct.

- Example 2

Perform the following division of polynomials by the synthetic division method

(7x ³ -x + 2): (x + 2)

Solution

In this case we have that the term x ² does not appear, so we will write 0 as its coefficient. Thus, the polynomial would be 7x ³ + 0x ² -x + 2.

We write their coefficients in a row, this is:

We write the value of C = -2 on the left side of the second row and draw the division lines.

We lower the leading coefficient b _n-1 = 7 and multiply it by -2, writing its result in the second row to the right.

We add and proceed as previously explained, until we reach the last term:

In this case, the remainder is r (x) = - 52 and the quotient obtained is q (x) = 7x ² -14x + 27.

- Example 3

Another way to use synthetic division is the following: suppose we have a polynomial P (x) of degree n and we want to know what value is by evaluating it at x = c.

By the division algorithm we can write the polynomial P (x) in the following way:

In this expression q (x) and r (x) are the quotient and the remainder, respectively. Now, if d (x) = x- c, when evaluating at c in the polynomial we get the following:

Therefore, it only remains to find ar (x), and we can do this thanks to the synthetic division.

For example, we have the polynomial P (x) = x ⁷ -9x ⁶ + 19x ⁵ + 12x ⁴ -3x ³ + 19x ² -37x-37 and we want to know what its value is by evaluating it at x = 5. To do this we perform the division between P (x) and d (x) = x -5 by the synthetic division method:

Once the operations are done, we know that we can write P (x) in the following way:

P (x) = (x ⁶ -4x ⁵ –x ⁴ + 7x ³ + 32x ² + 179x + 858) * (x-5) + 4253

Therefore, when evaluating it we have to:

P (5) = (5-4 (5) -5 + 7 (5) +32 (5) +179 (5) +858) * (5-5) + 4253

P (5) = (5-4 (5) -5 + 7 (5) +32 (5) +179 (5) +858) * (0) + 4253

P (5) = 0 + 4253 = 4253

As we can see, it is possible to use synthetic division to find the value of a polynomial by evaluating it at c rather than simply substituting c for x.

If we tried to evaluate P (5) in the traditional way, we would be forced to perform some calculations that often become tedious.

- Example 4

The division algorithm for polynomials is also true for polynomials with complex coefficients and, as a consequence, we have that the synthetic division method also works for such polynomials. We will see an example below.

We will use the synthetic division method to show that z = 1+ 2i is a zero of the polynomial P (x) = x ³ + (1 + i) x ² - (1 + 2i) x + (15 + 5i); that is, the remainder of the division P (x) by d (x) = x - z is equal to zero.

We proceed as before: in the first row we write the coefficients of P (x), then in the second we write z and draw the division lines.

We carry out the division as before; this is:

We can observe that the remainder is zero; therefore we conclude that z = 1+ 2i is a zero of P (x).

References

1. Baldor Aurelio. Algebra Grupo Editorial Patria.
2. Demana, Waits, Foley & Kennedy. Precalculus: Graphical, Numerical, Algebraic 7th Ed. Pearson Education.
3. Flemming W & Varserg D. Algebra and Trigonometry with Analytical Geometry. Prentice hall
4. Michael Sullivan. Precalculus 4th Ed. Pearson Education.
5. Red. Armando O. Algebra 1 6th Ed. The Athenaeum.