# Synthetic Division Method

I must say that synthetic division is the most “fun” way of dividing polynomials.  It has fewer steps to arrive at the answer as compared to the polynomial long division method. In this lesson, I will go over five (5) examples that should hopefully make you familiar with the basic procedures in successfully dividing polynomials using synthetic division.

## Things to Remember:

• Make sure the dividend is in standard form. That means the powers are in decreasing order.
• The divisor must be in the form x - \left( c \right).

## Examples of How to Divide Polynomials Using the Synthetic Division

Example 1: Divide the polynomial below.

Let us re-examine the given problem and make the necessary adjustments, if necessary.

The dividend (stuff to divide) is in standard form because the exponents are in decreasing order. That’s good!

The divisor needs to be rewritten as

At this point, I can now set up the synthetic division by extracting the coefficients of the dividend and then lining them up on top.

Directly to the left side, place the value of c = - 2 inside the “box”.

Finally, construct a horizontal line just below the coefficients of the dividend.

Steps:

1. Drop the first coefficient below the horizontal line.

2. Multiply that number you drop by the number in the “box”. Whatever its product, place it above the horizontal line just below the second coefficient.

3. Add the column of numbers, then put the sum directly below the horizontal line.

4. Repeat the process until you run out of columns to add.

See the animated solution below:

The last number below the horizontal line is always the remainder! The remainder of this problem is 3.

So how do we present our final answer?

Notice that the numbers below the horizontal line except the last (remainder) are the coefficients of the Quotient.

More so, the exponents of the variables of the quotient are all reduced by 1.

Example 2: Divide the polynomial.

This is not a trick question. Notice that the quotient does not have all the exponents of the variable x.

I can see that we are missing {x^4} and {x^2}. To include all the coefficients of variable x in decreasing power, we should rewrite the original problem like this. Attach zeroes on those missing x‘s. Also express the divisor as x - (c) which clearly reveals the value of c, that is, c = + 1.

From this point, I can now set up the numbers to continue with the process.

Steps:

1. Drop the first coefficient below the horizontal line.

2. Multiply that number you drop by the number in the “box”. Whatever its product, place it above the horizontal line just below the second coefficient.

3. Add the column of numbers, then put the sum directly below the horizontal line.

4. Repeat the process until you run out of columns to add.

See animated solution below:

So putting the final answer in the form

we have

Example 3: Divide the polynomial below.

\left( { - 2{x^4} + x} \right) \div \left( {x - 3} \right)

This is becoming more interesting! The quotient definitely looks horrible because it is missing a lot. Not only it lacks some x‘s which are {x^3} and {x^2} but the constant is also gone.

To fix this, I will rewrite the original problem in such a way that all x‘s are accounted for. But more importantly, do not forget to include the missing constant which is zero.

The “new and improved” problem should look like this:

From here, proceed with the steps as usual.

Steps:

1. Drop the first coefficient below the horizontal line.

2. Multiply that number you drop by the number in the “box”. Whatever its product, place it above the horizontal line just below the second coefficient.

3. Add the column of numbers, then put the sum directly below the horizontal line.

4. Repeat the process until you run out of columns to add.

See the animated solution below:

Okay then, the final answer for this is

You can write the final answer in two ways. The first one is using the minus or subtraction symbol to indicate that the remainder is negative. The second one is using the + symbol but attaching a negative symbol to the numerator. They mean the same thing!

Example 4: Divide the polynomial below.

\left( { - {x^5} + 1} \right) \div \left( {x + 1} \right)

Don’t be discouraged by this problem. This is actually quite easy especially now that you have gone through a few examples already. Always remember to “fill in the missing parts”, right?

Observe the dividend and you should agree that the missing parts are {x^4}, {x^3}, {x^2}, and x.

Rewriting the original problem that is synthetic-division ready, we get…

We populated the missing x‘s with zeros and explicitly solve for c = -1.

Steps:

1. Drop the first coefficient below the horizontal line.

2. Multiply that number you drop by the number in the “box”. Whatever its product, place it above the horizontal line just below the second coefficient.

3. Add the column of numbers, then put the sum directly below the horizontal line.

4. Repeat the process until you run out of columns to add.

See the animated solution below:

The last number below the horizontal line will always be the remainder. Don’t forget that. In this case, the remainder equals 2.

Example 5: Divide the polynomial by a binomial.

In this example, we will get a remainder of zero. When that happens the divisor becomes a factor of the dividend. In other words, the divisor evenly divides the dividend.

By examining the problem, I see that there are no missing components. All powers of x‘s are accounted for, and we have a constant. That’s great! This problem is in fact synthetic-division ready.

Steps:

1. Drop the first coefficient below the horizontal line.

2. Multiply that number you drop by the number in the “box”. Whatever its product, place it above the horizontal line just below the second coefficient.

3. Add the column of numbers, then put the sum directly below the horizontal line.

4. Repeat the process until you run out of columns to add.

See animated solution below:

Because the remainder equals zero, this means the divisor x - 5 is a factor of the dividend

therefore

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