Synthetic Division
2026-02-28 23:45 Diff
https://brightchamps.com/en-us/math/algebra/synthetic-division

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:


\({p(x)\over q(x)} = {Q(x) + {R\over q(x)}} \)
\({{2x^3+5x^2-3x+4\over x-1}} = (2x^2 + 7x + 4) + {{8\over (x - 1)}} \)