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185 LearnersLast updated on October 21, 2025

Dividing polynomials quickly and easily is done using synthetic division. Normally, we use the long division method to divide polynomials, but it is lengthy and difficult to follow. Synthetic division is a shortcut. This article discusses synthetic division and the steps involved.
We use synthetic division instead of the long division method because it is faster, and we don’t have to write each variable and its exponents repeatedly. Since we only work with coefficients in synthetic division the process becomes easier. When we divide one polynomial by another, we can write it as:
\({p(x)\over q(x)} = {Q(x) + {R\over q(x)}}\)
Here, p(x) is the dividend.
q(x) is the divisor.
Q(x) is the quotient we get from the division.
R is the remainder.
The key differences between synthetic division and long division are summarized in the table below
|
Aspect |
Synthetic Division |
Long Division |
|
What’s it for |
Synthetic division is the fastest way to divide when the divisor looks like (x-number).
|
Long division is used to divide any kind of polynomial. |
|
What you need |
The divisor must be simple, like \((x - 2)\).
|
Long division can be any polynomial, like \((x^2 + x + 2)\). |
|
How it works |
Uses only the numbers, not the variables.
|
You have to manage variables and their powers during division. |
|
Speed |
It is much faster for simple divisors.
|
It takes more time and requires more writing. |
|
Number of steps |
It involves fewer steps and relies primarily on addition and multiplication.
|
More steps, divide, multiply, subtract, and bring the terms down. |
|
Best for |
When dividing by (x - number).
|
Best for any type of polynomial. |
Let’s see the step-by-step method for synthetic division using an example.
Divide \((2x^3 + 5x^2 - 3x + 4) {\text { by }} (x - 1) \)
Step 1: Check the polynomial
Check whether the given polynomial is in standard form. All terms should be arranged in descending order of powers.
The given polynomial is \(2x^3 + 5x^2 - 3x + 4\). Now write down only the coefficients \(2x^3 + 5x^2 - 3x + 4 \) becomes 2, 5, -3, 4.
And the given divisor is \((x - 1)\), we solve \(x - 1= 0\) and get \(x = 1\). This makes dividing easier.
Step 2: Set the synthetic division box
We have to create a division box with the divisor and the dividend.
Divisor: 1
Dividend: 2, 5, -3, 4
\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ \end{array} \)
Step 3: Bring down the first number
Just bring down the first number
\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ &\downarrow\\ & &2 \end{array} \)
Step 4: Multiply and add
Multiply the first number that we brought down by the divisor and write the answer under the next number.
\(1 × 2 = 2\)
Now we have to write the 2 below the next number that is 5, and add both numbers together.
\(5 + 2 = 7\)
Now do the same multiplication and addition with the number 7
\(1 × 7 = 7\\
-3 + 7 = 4\)
And again, repeat the steps with 4
\(1 × 4 = 4\\
4 + 4 = 8\)
We get the number 8, which is the remainder, and the first three numbers will be the quotient.
\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ &\downarrow &2 & 7 & 4 \\ & &2 & 7 & 4 & 8 \\ \end{array} \)
Step 5: Write the final answer
The numbers at the bottom, except the last number, are quotient coefficients. We started the polynomial with a degree of 3, now we have to go one power lower.
Therefore, the quotient becomes
\(2x^2 + 7x + 4\)
The last number, 8, is the remainder.
The final answer should be in the format of:
\({p(x)\over q(x)} = {Q(x) + {R\over q(x)}}
\)
\({{2x^3+5x^2-3x+4\over x-1}} = (2x^2 + 7x + 4) + {{8\over (x - 1)}}
\)
Synthetic division simplifies dividing a polynomial by a linear binomial like \((x - 20)\). Here are some advantages and disadvantages of synthetic division.
Advantages of Synthetic Division
Disadvantages of Synthetic Division
Synthetic division is a quick and effect method to divide polynomials. In this section, we will learn some tips and tricks to make synthetic division.
Kids often make mistakes when doing synthetic division. Here are some of those mistakes and ways to avoid them, which can help them fix and understand them.
Synthetic division is not only used in math, but it also plays a major role in several real-life applications, especially in fields like engineering, healthcare, etc. Here are some of the applications:
Use synthetic division to divide: f(x) = 2x3 + 3x2 - 2x + 4 by x - 1.
Quotient: \(2x^2 + 5x + 3\)
Remainder: 7
1. The coefficients are 2, 3, -2, 4.
2. The divisor is x - 1, i.e. x = 1.
3. Using the synthetic division method, divide them:
\(
\begin{array}{r|rrrr}
& & & & \\
\hline
& & & & \\
1 & 2 & 3 & -2 & 4 \\
& &2 & 5 & 3
& & & & \\
\hline
& & & & \\
& 2 & 5 & 3 & 7
\end{array}
\)
4. Write the quotient and the remainder. When writing the quotient, we have to reduce it to one power, and the last number is the remainder.
Quotient: \(2x^2 + 5x + 3\)
Remainder: 7
Use synthetic division to divide f(x) = 3x^3 - 5x + 2 by x + 2.
Quotient: \(3x^2 -6x + 7\)
Remainder: -12
1. Write the coefficients 3, 0, -5, 2. Here, the x2 term is missing, so we add 0 in its stead.
2. The given divisor is \(x + 2\), so the divisor becomes -2.
\(
\begin{array}{r|rrrr}
& & & & \\
\hline
& & & & \\
-2 & 3 & 0 & -5 & 2 \\
& &-6 & 12 & -14
& & & & \\
\hline
& & & & \\
& 3 & -6 & 7 & -12
\end{array}
\)
Quotient: \(3x^2 -6x + 7\)
Remainder: -12
Divide 5x^2 - 3x + 7 by x + 1.
Quotient: \(5x - 8\)
Remainder: 15
1. The coefficients are 5, -3, 7.
2. Dividing x + 1, so the divisor will become -1.
\(
\begin{array}{r|rrrr}
& & & & \\
\hline
& & & & \\
-1 & 5 & -3 & 7\\
& &-5 & 8
& & & & \\
\hline
& & & & \\
& 5 & -8 & 15
\end{array}
\)
Quotient: 5x - 8
Remainder: 15
Divide 2x^3 + 7x^2 - x + 5 by x - 3
Quotient: \(2x^2 + 13x + 38\)
Remainder: 119
1. The coefficients are 2, 7, -1, 5.
2. The divisor is 3.
\(
\begin{array}{r|rrrr}
& & & & \\
\hline
& & & & \\
3 & 2 & 7 & -1 & 5\\
& &6 & 39 & 114
& & & & \\
\hline
& & & & \\
& 2 & 13 & 38 & 119
\end{array}
\)
Quotient: \(2x^2 + 13x + 38\)
Remainder: 119
Divide x^2 + 2x + 1 by x + 1
Quotient: \(x + 1\)
Remainder: 0
1. Coefficients: 1, 2, 1.
2. Divisor: -1
\(
\begin{array}{r|rrrr}
& & & & \\
\hline
& & & & \\
-1 & 1& 2 & 1 \\
& &-1 & -1
& & & & \\
\hline
& & & & \\
& 1 & 1 & 0
\end{array}
\)
Quotient: x + 1
Remainder: 0
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






