Math (SAT®)43 views·Updated Aug 17, 2026·3 pages

Mastering Synthetic Division of Polynomials

Natalie Fritz@atalieritz_kziv

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Dividing polynomials might seem challenging at first, but it's actually...

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Synthetic Division of Polynomials – page 1

Long Division of Polynomials

Just like dividing regular numbers, polynomial long division helps you split one polynomial by another. The goal is to find a quotient and remainder that satisfy: $\frac{f(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}$ . Always remember that the remainder's degree must be less than the divisor's degree.

When setting up polynomial long division, organize it like regular long division but leave placeholders (zeros) for any missing terms. For example, when dividing $16x^3+8x^2+29x+7$ by $4x+1$ , you start by dividing the first terms ( $16x^3 ÷ 4x = 4x^2$ ) and continue the process systematically.

Through the division process, you multiply, subtract, bring down terms, and repeat until the remainder's degree is less than the divisor's. In our example, the final answer is $4x^2+x+7+\frac{9}{4x+1}$ , showing both the quotient and remainder parts.

Try This! When dividing polynomials, make sure to line up terms with the same powers of x. This organization makes the subtraction steps much easier and helps prevent errors.

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Synthetic Division of Polynomials

Synthetic division is a shortcut method that works when dividing any polynomial by a linear factor in the form $(x-k)$ . This technique uses a compact arrangement of just the coefficients, making the process much faster than long division.

To set up synthetic division, place the value of $k$ (from the divisor $x-k$ ) on the left of your work. Write all the coefficients of the dividend in a row (including zeros for missing terms). Then follow the pattern of bringing down the first coefficient, multiplying by $k$ , adding down the column, and repeating.

The Remainder Theorem tells us that if a polynomial $f(x)$ is divided by $x-k$ , then $f(k)$ equals the remainder. Related to this, the Factor Theorem states that $x-k$ is a factor of polynomial $f(x)$ if and only if $f(k) = 0$ . This gives us a quick way to check if something is a factor!

Important Connection: When the remainder equals zero in synthetic division, the divisor $(x-k)$ is definitely a factor of the original polynomial. This also means that $k$ is a solution to the polynomial equation.

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Synthetic Division of Polynomials – page 3

More Synthetic Division Examples

Synthetic division saves time, especially with more complex polynomials. For example, when dividing $2x^3-15x^2+21x+24$ by $x-3$ , we set up the synthetic division with $k=3$ and work through the process. The result shows a quotient of $2x^2-9x-6$ with a remainder of $6$ , written as $2x^2-9x-6+\frac{6}{x-3}$ .

When using synthetic division, always remember to include zeros for any missing terms in your polynomial. For instance, in $2x^4-7x^2-4x+2$ divided by $x+1$ , we need to include a zero for the missing $x^3$ term before starting the division process.

The final answer format depends on whether the divisor is a factor. If the remainder is zero (as in the fifth example where $3x^4-25x^3+27x^2+6x+7$ divided by $x-7$ gives a remainder of 0), then the divisor is a factor and the answer is just the quotient. Otherwise, write the answer with the remainder over the divisor.

Pro Tip: When dividing by $(x+a)$ instead of $(x-k)$ , use $k=-a$ in your synthetic division. For example, to divide by $(x+1)$ , use $k=-1$ in your setup.

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Math (SAT®)43 views·Updated Aug 17, 2026·3 pages

Mastering Synthetic Division of Polynomials

Natalie Fritz@atalieritz_kziv

Add to Google sources

Dividing polynomials might seem challenging at first, but it's actually a systematic process that follows clear steps. In these notes, we'll explore two methods for dividing polynomials: long division and synthetic division, which give you powerful tools to solve polynomial...

of 3

Long Division of Polynomials

Try This! When dividing polynomials, make sure to line up terms with the same powers of x. This organization makes the subtraction steps much easier and helps prevent errors.

of 3

Synthetic Division of Polynomials

Important Connection: When the remainder equals zero in synthetic division, the divisor $(x-k)$ is definitely a factor of the original polynomial. This also means that $k$ is a solution to the polynomial equation.

of 3

More Synthetic Division Examples

Pro Tip: When dividing by $(x+a)$ instead of $(x-k)$ , use $k=-a$ in your synthetic division. For example, to divide by $(x+1)$ , use $k=-1$ in your setup.

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Long Division of Polynomials

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More Synthetic Division Examples

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