Polynomial Long Division Lesson

Over the course of the last few lessons, we have learned how to add, subtract, and multiply polynomials. Our next step is to learn how to divide polynomials.

Dividing a Polynomial by a Monomial

We will begin with the easiest scenario, which is to divide a polynomial by a monomial. In our pre-algebra course, we learned the following property: $$\frac{a + b}{c}=\frac{a}{c}+ \frac{b}{c}$$ As an example: $$\frac{5 + 10}{5}=\frac{5}{5}+ \frac{10}{5}$$ Either way, the result is 3. $$\frac{5 + 10}{5}=\frac{15}{5}=3$$ $$\frac{5}{5}+ \frac{10}{5}=1 + 2=3$$ We can use the same property when dividing a polynomial by a monomial. To divide a polynomial by a monomial, we divide each term of the polynomial (dividend) by the monomial (divisor). Let's look at a few examples.
Example 1: Find each quotient. $$(2x^3+16x^2+20x) ÷ (4x)$$ Let's set this problem up as a fraction: $$\frac{2x^3+16x^2+20x}{4x}$$ We will divide each term of the dividend, by the divisor. $$\frac{2x^3}{4x}+ \frac{16x^2}{4x}+ \frac{20x}{4x}$$ Simplify each part. $$\frac{x^2}{2}+ 4x + 5$$ Example 2: Find each quotient. $$(2x^3 + 10x^2 + 2x) ÷ (10x^3)$$ Let's set this problem up as a fraction: $$\frac{2x^3 + 10x^2 + 2x}{10x^3}$$ We will divide each term of the dividend, by the divisor. $$\frac{2x^3}{10x^3}+ \frac{10x^2}{10x^3}+ \frac{2x}{10x^3}$$ Simplify each part. $$\frac{1}{5}+ \frac{1}{x}+ \frac{1}{5x^2}$$

Polynomial Long Division

We can divide a polynomial by a non-monomial using polynomial long division. This process is similar to long division with whole numbers. Let's take a look at an example.
Example 3: Find each quotient. $$(x^3 - 7x^2 + 5x + 28) ÷ (x - 4)$$ Our first step is to make sure each polynomial is in standard form. In this case, both our dividend and divisor are in standard form. The next step is to set up a long division. Just like when we performed long division with whole numbers, the dividend goes under the long division symbol and the divisor goes out to the left of the long division symbol. Now we will move through the long division process using the DMSBR method.
D » Divide
M » Multiply
S » Subtract
B » Bring Down
R » Repeat or Remainder
For the division step, we divide the leading term of the dividend (x³) by the leading term of the divisor (x). The result (x²) is written above the (-7x²): For the multiplication step, we multiply our result (x²) by the divisor (x - 4). We write the result (x³ - 4x²) directly below the (x³ - 7x²): For the subtraction step, we subtract away our result (x³ - 4x²) from the multiplication step. To perform this action, we will change the sign of each term into its opposite. Once this is done, we can add: For the bring down step, we bring the next term down from our dividend. In this case, we will bring down a 5x. For the last step in the series, we see repeat or remainder. If we brought down a term in the last step, this tells us to repeat the process. We will go back to the division step.
For the second division step, we are now working with the bottom polynomial (-3x² + 5x). We will divide the leading term (-3x²) by the leading term of the divisor (x). The result (-3x) will be placed above 5x: For the second multiplication step, we multiply our result (-3x) by the divisor (x - 4). We write the result (-3x² + 12x) directly below the (-3x² + 5x): For the second subtraction step, we subtract away our result (-3x² + 12x) from the multiplication step. To perform this action, we will change the sign of each term into its opposite. Once this is done, we can add: For the second bring down step, we bring down the next term from our dividend. In this case, we will bring down a 28. For the second repeat or remainder step, we repeat our process since we brought down a term in the last step.
For the third division step, we are now working with the bottom polynomial (-7x + 28). We will divide the leading term (-7x) by the leading term of the divisor (x). The result (-7) will be placed above 28: For the third multiplication step, we multiply our result (-7) by the divisor (x - 4). We write the result (-7x + 28) directly below the (-7x + 28): For the third subtraction step, we subtract away our result (-7x + 28) from the multiplication step. To perform this action, we will change the sign of each term into its opposite. Once this is done, we can add: Since there are no more terms to bring down, we move into our repeat or remainder step. In this case, our final subtraction results in a zero. When this occurs, we do not have a remainder.
$$(x^3 - 7x^2 + 5x + 28) ÷ (x - 4)=x^2 - 3x - 7$$ We can always check the result of division using multiplication. If we multiply our quotient by the divisor, we should get our dividend back: $$(x-4)(x^2 - 3x - 7)=x^3 - 7x^2 + 5x + 28$$ Let's look at another example.
Example 4: Find each quotient. $$(-29x + 35x^3 - 51x^2 - 11) ÷ (7x + 1)$$ $$(35x^3-51x^2-29x-11) ÷ (7x + 1)=5x^2 - 8x - 3 - \frac{8}{7x+1}$$ The remainder of -8 is written over the divisor (7x + 1). This is added to our quotient (5x² - 8x - 3).

Dividing Polynomials with Missing Powers of the Variable

In some cases, we will divide polynomials that have missing powers of the variable. When this happens, we write a 0 as the coefficient for each missing power. Let's take a look at an example.
Example 5: Find each quotient. $$(x^3 + 2x - 3) ÷ (x - 1)$$ If we look at our dividend, we see that we are missing x². We can rewrite our problem as: $$(x^3 + 0x^2 + 2x - 3) ÷ (x - 1)$$ $$(x^3 + 0x^2 + 2x - 3) ÷ (x - 1)=x^2 + x + 3$$

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