# Synthetic Division (With Steps & Examples)

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There are several ways to divide a polynomial by another. The most basic way to do it is by using long division. But long division is very lengthy and time consuming. Especially for higher degree polynomials, long division take pretty long time to get a quotient and remainder. Synthetic Division is easy, fast and very straightforward. It helps students save a lot of time when compared with other traditional methods.

## What is a Polynomial?

A polynomial is an algebraic expression made up of two or more terms subtracted, added, or multiplied. A polynomial can contain coefficients, variables, exponents, constants, and operators such as addition and subtraction.

It is also important to note that a polynomial can’t have fractional or negative exponents.

Examples of polynomials are: 4y2 – 2y + 7, -2x3 + 2x2 − 7x + 8, (¾)x2 – 7x + 3 etc. Like numbers, polynomials can undergo addition, subtraction, multiplication, and division.
Examples of algebraic expressions that are not polynomials are: 5x4/5 + 7x + 3, 2x-2 + 7x – 3. (Having fractional or negative numbers as indices of a variable).

## Division of Polynomials

Dividing polynomials is an algorithm to solve a rational number that represents a polynomial divided by a monomial or another polynomial. The divisor and the dividend are placed exactly the same way as we do for regular division. For example, if we need to divide 6x2 – 5x + 18 by 3x + 7, we write it in this way:

The polynomial written on top of the bar is the numerator ( 6x2 – 5x + 18), while the polynomial written below the bar is the denominator (3x + 7). This can be understood by the following figure which shows that the numerator becomes the dividend and the denominator becomes the divisor.

## What is the Synthetic Division?

Synthetic division is a shortcut method of dividing a polynomial by a linear polynomial (polynomial of degree 1). It is a simplified way of finding the zeroes of polynomials.

One of the advantages of using this method over the traditional long method is that synthetic division is performed manually with less effort. Since synthetic division allows one to calculate without writing the variables while performing the polynomial division, it reduces the chances of making mistakes (the ones that students make while solving a problem using a long division method).

In synthetic division, a linear binomial of the form (x – a) is used as a divisor. When  we divide a polynomial P(x) of degree n by a linear polynomial (x – a), we get a polynomial quotient Q(x) of degree (n – 1) and a constant polynomial R as the remainder.

Mathematically it can be represented as:

P(x)/(x – a) = Q(x) + R/(x – a)

Where, P(x) is a polynomial of degree n

(x – a) is a binomial of degree 1

Q(x) is a polynomial of degree (n – 1).

R is a constant polynomial, i.e., a number without variable

Hence, we can use the synthetic division method to find the remainder quickly.

## Steps to Perform Synthetic Division

Following steps are carried out to perform synthetic division on a polynomial P(x) of degree n and a binomial (x – a):

Step 1: Write a for the divisor.

Step 2: Write the coefficients of the dividend.

Step 3: Bring the leading coefficient down.

Step 4: Multiply the leading coefficient by a. Write the product in the next column.

Step 5: Add the terms of the second column.

Step 6: Multiply the result by a. Write the product in the next column.

Step 7: Repeat steps 5 and 6 for the remaining columns.

Step 8: Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on.

## Examples

Let’s take a few examples to understand the process.

### Example 1

Divide (-3x3 + 19x2 – 30x + 15) by (x – 4)

Note: (x – 4) is a linear polynomial (degree 1), so synthetic division is a fit case.

Step 1: P(x) = (-3x3 + 19x2 – 30x + 15)

(x – a) = (x – 4) => a = 4
Step 2: Coefficients of -3x3 + 19x2 – 30x + 15 are -3, 19, -30 and 15

Write a in the next line

Now, draw a horizontal line

Step 3: Leading coefficient is -3 and is brought down

Step 4: Multiply the leading coefficient -3 by a(= 4). Write the product in the next column

Step 5: Add the terms of the second column.

Step 6: Multiply the result by a(4). Write the product in the next column.

Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively.

Rightmost number, i.e., 7 is remainder

Now, for remaining numbers (-3, 7 and  -2) start from right:

Power of variable x for -2 is 0, i.e., x0

Power of variable x for -7 is 1, i.e., x1

Power of variable x for 3 is 2, i.e., x2

Therefore, quotient Q(x) = -3x2 + 7x – 2 and remainder R = 7

Let’s check the result.

Recall the division algorithm: Dividend = (Divisor × Quotient) + Remainder.

Divisor = (x – 4)

Quotient = (-3x2 + 7x – 2)

Remainder = 7

(Divisor × Quotient) + Remainder = (x – 4) × (-3x2 + 7x – 2) + 7

= x × (-3x2 + 7x – 2) + (-4) × (-3x2 + 7x – 2) + 7

= (x ×  (-3x2)+ x × 7x  + x × (- 2)) + (-4 × (-3x2) + (-4) × 7x + (-4) × (- 2)) + 7

= (-3x3 + 7x2  – 2x + 12x2 – 28x + 8) + 7

= -3x3 + 19x2  – 30x + 8 + 7= -3x3 + 19x2  – 30x + 15 is the same polynomial that we divide by (x – 4)

### Example 2

Let’s now divide a bit higher degree polynomial

Divide (x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2) by (x + 3)

Step 1: P(x) = (x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2)

(x – a) = (x + 3) => a = -3

Step 2: Coefficients of x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2 are 1, 1, 1, 22, 8, 14 and 2

Write a in the next line

Now, draw a horizontal line

Step 3: Leading coefficient is 1 and is brought down

Step 4: Multiply the leading coefficient 1 by a(= -3). Write the product in the next column

Step 5: Add the terms of the second column.

Step 6: Multiply the result by a(-3). Write the product in the next column.

Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively.

That finishes the process and the quotient is x5 – 2x4 + 7x3 + x2 + 5x – 1 and remainder is 5.

Check: (x + 3) × (x5 – 2x4 + 7x3 + x2 + 5x – 1) + 5 = x6 + x5 + x4 + 22x3 + 8x2 + 14x + 2

### Example 3

Divide (x4 – 6) by (x + 1)

Here you can see that the terms for x3, x2 and x are not there. In  all such cases, we add such terms with coefficient 0.

Step 1: (x4 – 6) is written as (x4 + 0x3 + 0x2 + x – 6).

And, (x + a) = (x + 1) => a = -1
Step 2: Coefficients of x4 + 0x3 + 0x2 + 0x – 6 are 1, 0, 0, 0 and -6

Write a in the next line

Now, draw a horizontal line

Step 3: Leading coefficient is 1 and is brought down

Step 4: Multiply the leading coefficient 1 by a(= -1). Write the product in the next column

Step 5: Add the terms of the second column.

Step 6: Multiply the result by a(-1). Write the product in the next column.

Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively.

That finishes the process and the quotient is x3 – x2 + x – 1 and remainder is -5.

Check: (x + 1) × (x3 – x2 + x – 1) – 5

= x4 – x3 + x2 – x + x3 – x2 + x – 1 – 5

= x4 – 6

### Example 4

Divide (4x6 – 9x4 + 2x3 + 3x2 – 8x – 12) by (2x + 3)

Step 1: (4x6 – 9x4 + 2x3 + 3x2 – 8x – 12) is written as (4x6 + 0x5 – 9x4 + 2x3 + 3x2 – 8x – 12).

(2x + 3) = 2(x + 3/2)

Now, in order to get the form (x + a), we ignore 2 in 2(x + 3/2)

And, (x + a) = (x + 3/2) => a = -3/2

Step 2: Coefficients of 4x6 + 0x5 – 9x4 + 2x3 + 3x2 – 8x – 12 are 4, 0, -9, 2, 3, -8 and -12

Write a in the next line

Now, draw a horizontal line

Step 3: Leading coefficient is 4 and is brought down

Step 4: Multiply the leading coefficient 1 by a(= -3/2). Write the product in the next column

Step 5: Add the terms of the second column.

Step 6: Multiply the result by a(-3/2). Write the product in the next column.

Step 7: Repeat steps 5 and 6 for the remaining columns, i.e., add and multiply progressively.

That finishes the process and the quotient is 4x5 – 6x4 + 2x2 – 8 and remainder is 0.

Since, in the beginning we ignored 2 in 2(x + 3/2),

so here also, ignore 2 in the result 4x5 – 6x4 + 2x2 – 8

= 2(2x5 – 3x4 + x2 – 4)

So, quotient is 2x5 – 3x4 + x2 – 4 and remainder is 0

Check: (2x + 3) × (2x5 – 3x4 + x2 – 4) = 4x6 – 6x5 + 2x3 – 8x + 6x5 – 9x4 + 3x2 – 12

= 4x6 – 9x4 + 2x3 + 3x2 – 8x – 12

## Advantages and Limitations of Synthetic Division

The advantages of using the synthetic division method are:

• The calculation can be performed without variables
• It requires only a few calculation steps
• Unlike the polynomial long division method, there is no step involving subtraction and hence this method is less error-prone.

The only limitation of the synthetic division method is that it is only applicable if the divisor is a linear polynomial (polynomial of degree 1).

## Conclusion

We hope you enjoyed our post on how to do synthetic division of polynomials. There are a variety of ways to divide polynomials, but synthetic division is the most quick and easiest to remember.