Divide Polynomials

Divide a Polynomial by a Monomial

In the last section you learned how to divide a monomial by a monomial. As you continue to build up your knowledge of polynomials, the next procedure is to divide a polynomial of two or more terms by a monomial.

The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start by reviewing fraction addition. The sum $\tfrac{y}{5} + \tfrac{2}{5}$ simplifies to $\tfrac{y+2}{5}$. Now we will do this in reverse to split a single fraction into separate fractions.

We’ll state the fraction addition property here just as you learned it, and in reverse.

We use the form on the left to add fractions, and we use the form on the right to divide a polynomial by a monomial. For example, $\tfrac{y+2}{5}$ can be written $\tfrac{y}{5} + \tfrac{2}{5}$. We use this form of fraction addition to divide polynomials by monomials.

Example. Find the quotient: $\tfrac{7y^{2}+21}{7}$.

Divide each term of the numerator by the denominator, then simplify each fraction:

Remember that division can be represented as a fraction. When you are asked to divide a polynomial by a monomial and it is not already in fraction form, write a fraction with the polynomial in the numerator and the monomial in the denominator.

Example. Find the quotient: $\left(18x^{3} - 36x^{2}\right) \div 6x$.

Rewrite as a fraction, divide each term of the numerator by the denominator, and simplify:

When we divide by a negative, we must be extra careful with the signs.

Example. Find the quotient: $\tfrac{12d^{2}-16d}{-4}$.

Divide each term of the numerator by the denominator, then simplify. Remember, subtracting a negative is like adding a positive:

Example. Find the quotient: $\tfrac{105y^{5}+75y^{3}}{5y^{2}}$.

Separate the terms, then simplify each one:

Example. Find the quotient: $\left(15x^{3}y - 35xy^{2}\right) \div (-5xy)$.

Rewrite as a fraction, separate the terms, and simplify:

Example. Find the quotient: $\tfrac{36x^{3}y^{2}+27x^{2}y^{2}-9x^{2}y^{3}}{9x^{2}y}$.

Separate the terms, then simplify each one:

Example. Find the quotient: $\tfrac{10x^{2}+5x-20}{5x}$.

Separate the terms, then simplify. Notice that the last term does not divide evenly, so it stays as a fraction:

Divide a Polynomial by a Binomial

To divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully at the steps we take when we divide a 3-digit number, $875$, by a 2-digit number, $25$.

We write the long division, divide the first two digits ($87$) by $25$ to get $3$, multiply $3$ times $25$ and write the product under the $87$, then subtract. We bring down the next digit and repeat the process. The completed long division looks like this:

We check division by multiplying the quotient by the divisor. If we did the division correctly, the product should equal the dividend: $35 \cdot 25 = 875$ ✓.

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

Example. Find the quotient: $\left(x^{2} + 9x + 20\right) \div (x+5)$.

Write it as a long division problem, being sure the dividend is in standard form. Divide the first term of the dividend, $x^{2}$, by the first term of the divisor, $x$, to get $x$; place it in the quotient over the $x$ term. Multiply $x$ times $x+5$ to get $x^{2}+5x$, line it up under the dividend, and subtract (it may be easier to change the signs and add). Then bring down the last term, $20$:

Now divide the first term of the remainder, $4x$, by $x$ to get $4$; place it in the quotient over the constant term. Multiply $4$ times $x+5$ to get $4x+20$, and subtract:

To check, multiply the quotient by the divisor: $(x+4)(x+5)$. You should get the dividend, $x^{2}+9x+20$ ✓. So the quotient is $x+4$.

When the divisor has a subtraction sign, we must be extra careful when we multiply the partial quotient and then subtract. It may be safer to show that we change the signs and then add.

Example. Find the quotient: $\left(2x^{2} - 5x - 3\right) \div (x-3)$.

Divide $2x^{2}$ by $x$ to get $2x$; place it in the quotient over the $x$ term. Multiply $2x$ times $x-3$ to get $2x^{2}-6x$. Subtract by changing the signs and adding, then bring down the last term:

Now divide $x$ by $x$ to get $1$; place it in the quotient over the constant term. Multiply $1$ times $x-3$ to get $x-3$, and subtract by changing the signs and adding:

To check, multiply $(x-3)(2x+1)$. The result should be $2x^{2}-5x-3$. So the quotient is $2x+1$.

When we divided $875$ by $25$, there was no remainder. But sometimes division of numbers does leave a remainder, and the same is true when we divide polynomials. We write the remainder as a fraction with the divisor as the denominator.

Example. Find the quotient: $\left(x^{3} - x^{2} + x + 4\right) \div (x+1)$.

Divide $x^{3}$ by $x$ to get $x^{2}$; multiply $x^{2}$ times $x+1$ and subtract. Bring down the next term, divide again to get $-2x$, and continue. Finally, dividing $3x$ by $x$ gives $3$; multiplying $3$ times $x+1$ gives $3x+3$, and subtracting leaves a remainder of $1$:

We write the remainder as a fraction with the divisor as the denominator, so the quotient is:

To check, multiply $(x+1)\left(x^{2}-2x+3+\tfrac{1}{x+1}\right)$. The result should be $x^{3}-x^{2}+x+4$.

Look back at the dividends in the last three examples. The terms were written in descending order of degrees, and there were no missing degrees. When a dividend is missing a degree, we add that term in with a zero coefficient as a placeholder so the columns line up during long division.

Example. Find the quotient: $\left(x^{4} - x^{2} + 5x - 2\right) \div (x+2)$.

Notice that there is no $x^{3}$ term in the dividend. We will add $0x^{3}$ as a placeholder. Then long-divide as usual, changing signs and adding at each subtraction step:

To check, multiply $(x+2)\left(x^{3}-2x^{2}+3x-1\right)$. The result should be $x^{4}-x^{2}+5x-2$. So the quotient is $x^{3}-2x^{2}+3x-1$.

In the next example we will divide by $2a-3$. As we divide, we will have to consider the constants as well as the variables.

Example. Find the quotient: $\left(8a^{3} + 27\right) \div (2a+3)$.

This time we will show the division all in one step. We need to add two placeholders, $0a^{2}$ and $0a$, in order to divide:

To check, multiply $(2a+3)\left(4a^{2}-6a+9\right)$. The result should be $8a^{3}+27$. So the quotient is $4a^{2}-6a+9$.

Key terms

divide a polynomial by a monomial — divide each term of the polynomial by the monomial, using the property $\tfrac{a+b}{c} = \tfrac{a}{c} + \tfrac{b}{c}$. polynomial long division — a procedure, modeled on long division of numbers, for dividing a polynomial by a binomial: divide, multiply, subtract (or change the signs and add), and bring down, repeating until the remainder has lower degree than the divisor. placeholder — a term with a zero coefficient (such as $0x^{3}$) inserted for a missing degree so the columns line up during long division. remainder — the leftover after long division, written as a fraction with the divisor as the denominator.

This section is adapted from Elementary Algebra 2e, Section 6.6: Divide Polynomials by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: condensed the multi-row worked-example step tables into prose with a single typeset result, and recreated the numeric and polynomial long divisions as typeset math arrays; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.