Polynomial Division by Factoring

There are many different methods for dividing polynomials, but one thing remains the same: every division will have a divisor (the denominator of the fraction) and the dividend (the numerator of the fraction). Also, in many cases, the same division problem can be solved in multiple ways. To highlight this fact, we're going to do the same problem three different ways, explaining each technique in general and using the same problem as a repeated example.

Our task will be to divide

$$\frac{x^3 + 12x^2 + 20x}{x^2 + 10x},$$

where we assume $x \neq 10$ (that way we aren't dividing by zero).

Method 1: Factorization

Sometimes, factoring the numerator and/or the denominator can make things a lot easier. Even if this doesn't make the whole thing simple, it will make the math easier to handle. But this trick doesn't always work, especially if the number you're dividing by doesn't go evenly into the number you're dividing. Still, when it does work, it works really well! That is why it is the first thing we try.

First notice that the numerator and the denominator have a common factor of $x.$

$$\frac{x^3 + 12x^2 + 20x}{x^2 + 10x} = \frac{x(x^2 + 12x + 20)}{x(x + 10)}.$$

Now we can $\color{#8925AE}{\text{CANCEL}}$ the factors of $x$, and we are left with

$$\frac{x^3 + 12x^2 + 20x}{x^2 + 10x} = \frac{\cancel{x}(x^2 + 12x + 20)}{\cancel{x}(x + 10)} = \frac{x^2 + 12x + 20}{x + 10}.$$

We are left with a quadratic in the numerator and a linear binomial in the denominator. It turns out we can factor the numerator again, as follows:

$$\frac{x^2 + 12x + 20}{x + 10} = \frac{(x + 10)(x + 2)}{x + 10}.$$

Now it is apparent the numerator and denominator have a common factor of $x + 10:$

$$\frac{x^2 + 12x + 20}{x + 10} = \frac{\cancel{(x + 10)}(x + 2)}{\cancel{x + 10}} = x + 2$$

Therefore our final answer is $\boxed{x + 2}.$

Now, we're going to see two more ways to solve this problem. It's a good idea to try different methods because it helps you check your work. If the new answer doesn't match the one you got before, there is likely a mistake.

Method 2: Yanking Factors

Perhaps you weren't able to recognize the factorization done in Method 1. While it's important to notice these factorization patterns, there are other methods that get around this. One of them involves pulling factors out of the numerator term-by-term (this is why I call it "Yanking Factors"). It is helpful if you can't seem to factor the numerator and/or denominator, but it will not help if the polynomial in the numerator has a lower degree than the polynomial in the denominator. In the fraction

$$\frac{x^3 + 12x^2 + 20x}{x^2 + 10x},$$

the numerator is a cubic and the denominator is a quadratic, so we're OK.

Here's how the method works: we pull off the last term in the numerator and make it its own, "new" fraction:

$$\frac{x^3 + 12x^2 + 20x}{x^2 + 10x} = \frac{x^3 + 12x^2}{x^2 + 10x} + \frac{20x}{x^2 + 10x}.$$

Now we borrow terms from the numerator of the "old" fraction so that we can cancel factors in the new fraction. $20x = 10x \times 2,$ so if we add $x^2 \times 2 = 2x^2$ to the second fraction, we will get a multiple of $x^2 + 10x$ in the denominator. It looks like this:

$$\begin{align} \frac{x^3 + 12x^2}{x^2 + 10x} + \frac{20x}{x^2 + 10x} &= \\ \frac{x^3 + 10x^2 + 2x^2}{x^2 + 10x} + \frac{20x}{x^2 + 10x} &= \\ \frac{x^3 + 10x^2}{x^2 + 10x} + \frac{2x^2}{x^2 + 10x} + \frac{20x}{x^2 + 10x} &= \\ \frac{x^3 + 10x^2}{x^2 + 10x} + \frac{2x^2 + 20x}{x^2 + 10x}. \end{align}$$

Now we've made some headway with the second fraction. The quadratic term in the numerator is twice the one in the denominator, and the same is true for the linear terms. That is, $2(x^2 + 10x) = 2x^2 + 20x,$ so in the second fraction we have a common factor of $x^2 + 10x.$

$$\begin{align} \frac{x^3 + 10x^2}{x^2 + 10x} + \frac{2x^2 + 20x}{x^2 + 10x} &= \\ \frac{x^3 + 10x^2}{x^2 + 10x} + \frac{2(x^2 + 10x)}{x^2 + 10x} &= \\ \frac{x^3 + 10x^2}{x^2 + 10x} + \frac{2\cancel{(x^2 + 10x)}}{\cancel{x^2 + 10x}} &= \\ \frac{x^3 + 10x^2}{x^2 + 10x} + 2. \end{align}$$

We still have a fraction, but the numerator is simpler. There's nothing more to be done than to try the same process again. The lowest-ordered term in the numerator is a factor of $x$ more than the lowest-ordered term in the denominator, so we want to build a fraction where there is also a term that is a factor of $x$ more than $x^2.$ With that in mind, we split our fraction into two, isolating the last term in the numerator:

$$\frac{x^3 + 10x^2}{x^2 + 10x} + 2 = \frac{x^3}{x^2 + 10x} + \frac{10x^2}{x^2 + 10x} + 2.$$

Since $x^2 \times x = x^3,$ we need to borrow an $x^3$ term from the first fraction to get a multiple of $x^2 + 10x$ in the numerator of the second fraction. But $x^3$ is the only term left in the numerator of the first fraction! So, there are no leftovers and we know that $x^3 + 10x^2$ has a factor of $x^2 + 10x.$ We also know that the other factor is $x$ because the lowest-ordered terms in the numerator and denominator differ by a factor of $x$ (that is, $10x \times x = 10x^2$). Therefore,

$$\frac{x^3 + 10x^2}{x^2 + 10x} + 2 = \frac{x\cancel{(x^2 + 10x)}}{\cancel{x^2 + 10x}} + 2 = \boxed{x + 2}.$$

There is no longer a fraction, so we are done. We also got the same answer as before, which is good.

**Method 3: Polynomial Long Division **

We've saved the most involved method for second to last because it works for any polynomial long division problem. Since Methods 1 and 2 are faster when they do work, it's worth trying those first, but if they don't work, long division comes to the rescue. Another way to do long division is method $4$ called synthetic division.

Polynomial long division works very similarly to long division for whole numbers. Each step in long division for whole numbers comes from one place value in the number being divided. Instead of place values, however, the polynomial division is broken into steps based on the terms of the polynomial, in decreasing order of their degree. As a result it is important that both the dividend and divisor are in standard form before attempting long division. It is also necessary to make sure that the dividend has zero terms added. In this case, there is no constant term, so we pretend there is constant term with coefficient $0:$

$$x^3 + 12x^2 + 20x = x^3 + 12x^2 + 20x + \color{#8925AE}{0}.$$

Just like in long division of whole numbers, we write long division of polynomials by putting the dividend under the long division sign and the divisor to its left:

Now the first step is somewhat inspired by Method 2 in that we look at one pair of terms from the numerator and denominator at a time. Divide the first term in the numerator by the first term in the denominator, which gives $\frac{x^3}{x^2} = x.$ We indicate this by putting an $x$ above the long division sign:

Now we take the term we just added above the long division sign and multiply it by ALL terms to the left of the long division sign. $x(x^2 + 10x) = x^3 + 10x^2.$ We put this result on a new line underneath the dividend:

Notice that we aligned the terms so that the terms of the same degree are in the same column. This makes it much easier to keep organized and I recommend you do the same. Now we subtract these two quantities (which is just combining like terms):

We dropped the $20x$ down from the first row below the long division sign to the bottom row. This is just a reminder of what terms haven't been subtracted away from the dividend yet. Since we haven't run out of these terms, we repeat the first step again. Now, however, our first term of the dividend is $2x^2,$ so dividing this by the first term in the divisor gives $\frac{2x^2}{x^2} = 2.$ We place a $2$ above the long division sign:

Now we multiply the $2$ across all terms in $x^2 + 10x,$ placing the result at the bottom.

Finally, we subtract one last time:

We have run out of terms in the dividend and got a $0$ at the bottom. That means the fraction divides evenly, and there is no remainder. We found that

$$\frac{x^3 + 12x^2 + 20x}{x^2 + 10x} = x + 2.$$

Again, we got the same answer. We should feel really confident now.

If we run out of terms in the dividend and the result at the bottom is something other than $0,$ then that is the remainder. If we have a remainder, we add it to our final answer, divided by the original divisor. For instance, if the remainder were a $4,$ then we'd add $\frac{4}{x^2 + 10x}$ to our answer above. As a quick check, the remainder will always be a polynomial of lower degree than the divisor. So if we got a quadratic remainder, for instance, then we made a mistake.

In general, polynomial division problems can only be solved with Method 1 if Method 3 gives no remainder. Methods 2 and 3 will work regardless of whether there is a remainder or not. Method 3 is the most straightforward to execute but also takes the longest. Keep this in mind when deciding which method(s) to try.

If that breakdown wasn't enough or you want more examples, you can see more examples of long division and a summary of all the different techniques for dividing polynomials.

Dividing Polynomials

When you're dividing a polynomial by another term (whether it's a monomial, a binomial, a trinomial, or another polynomial), the key things to remember are to:

1. Find common factors between the numerator and denominator
2. Cancel out common factors

(The end of this explanation includes a step-by-step on how to use long division to divide a polynomial. If you're just interested in that, skip on down!)

$$\begin{array}{cccl} \color{#8925AE}{\text{Monomials}}&\color{#8925AE}{\text{Binomials}}&\color{#8925AE}{\text{Trinomials}}&\color{#8925AE}{\text{Polynomials}}\\ \color{#8925AE}{\text{(one term)}}&\color{#8925AE}{\text{(two terms)}}&\color{#8925AE}{\text{(three terms)}}&\color{#8925AE}{(\ge\text{2 terms})}\\ 3x & x+6 & x^2+3x-2 & x^3-3x^2\\ x & 2x^2+5x & 2x^3-x+5 & -x^4+6x^2+x+6\\ -2x & -x^3+7x & 4x^3+x^2+1 & x+x^2+x^3+x^4+x^{2019} \end{array}$$

As a refresher, here is a quick review on different kinds of polynomials:

Let's look at some examples, so we can see how to divide polynomials by monomials, binomials, and trinomials. Remember, our goals are to find and cancel out common factors.

(If you would like an extra resource, the WikiHow page linked in Stefan's explanation is a good place to start).

$$\frac{15x^3-72x^2-15x}{3x}$$

EXAMPLE 1: Dividing a Polynomial by a Monomial

It is always a good idea to see if you can factor out the denominator from the numerator.

$$\frac{3x(5x^2-24x-5)}{3x}$$

Now that we have factored out $3x$, we can cancel it out (we won't always be this lucky, however). This leaves us with a regular quadratic equation.

$$\frac{\cancel{3x}(5x^2-24x-5)}{\cancel{3x}}$$

So, after finding a common factor and canceling it out, and then factoring the remaining quadratic equation, we're left with:

$$\frac{3x(5x^2-24x-5)}{3x} = (5x+1)(x-5)$$

EXAMPLE 2: Dividing a Polynomial by a Binomial

$$\frac{4x^2-10x-24}{x-4}$$

Just like we did when dividing by a monomial, it is always a good idea to see if you can factor out the denominator from the numerator.

First, let's see if we can simplify the numerator.

After simplifying the numerator, we can rewrite our equation:

$$\frac{2(2x^2-5x-12)}{x-4}$$

Now, let's see if we can factor the denominator out of the numerator.

Let's rewrite the equation:

$$\frac{2(x-4)(2x+3)}{x-4}$$

$$\frac{2\cancel{(x-4)}(2x+3)}{\cancel{x-4}} = 2(2x+3)$$

EXAMPLE 3: Using Long Division

$$\frac{2x^2+7x-15}{x+5}$$

Let's say you are having trouble factoring the numerator, or maybe it is not able to be "cleanly" factored. A useful method is long division, which will work regardless of whether the numerator can be factored or not.

We can set up the long division just like we normally would, putting the divisor (the denominator) on the outside:

Here's what we need to do:

1. Divide the first term inside the division symbol by the first term outside it.

So, let's do this: divide $2x^{2}$ by $x$. We will write the result on top of the long division symbol (just like you normally would).

Next:

1. Multiply that resulting term by the whole expression outside the long division sign. (While that might sound confusing, you're actually just doing what you would always do with long division, just like if using long division on whole numbers).

2. Write the product under the term under the long division sign. (Again, this is just like what you would normally do).

Next:

1. Subtract the term you just wrote from the expression above it.

Next:

1. Bring down the remaining term from under the long division sign. Repeat the process.

Once you end up with $0$ or something that cannot be divided by the first term outside the division sign, you are done!

The results show us that:

$$\frac{2x^2+7x-15}{x+5} = 2x-3$$

(Video) Divide a Polynomial by a Monomial

by YourMathGal

The above video is from the YouTube channel YourMathGal, in which she describs how to divide polynomials in the case when the divisor is a monomial.

$\large{\text{Summary}}$

In the case in which the divisor is a binomial it is necessary to use the methods of long division or synthetic division.

Before applying any of these methods for any type of divisor, it might be useful to factor the numerator (and/or the denominator) if possible.

Polynomial Division by Factoring

Welcome to the first lesson on polynomial division! In this lesson, we will explore how to divide polynomials using factoring, a method that simplifies expressions before division.

Polynomial division by factoring works like simplifying fractions - you factor both the numerator and the denominator, then cancel out the common terms.

This method is a great way to break down complex expressions into simpler parts, especially when you're not ready to use long or synthetic division (which we'll cover in the next lessons!).

Let's look at a simple exam:

$$\frac{x^2-9}{x+3}$$

First, factor the numerator. We know that $x^2-9$ is a difference of squares, so it factors into: $(x-3)(x+3)$. Now our expression looks like:

$$\frac{(x-3)(x+3)}{x+3}$$

We see that $x+3$ appears in both the numerator and the denominator. So, we can cancel it out. We are left with:

$$x-3$$

This is your simplified answer!

In this method, it's important to factor completely and only cancel out the common factors, not terms. Also, make sure to check for any restrictions. In the above example, the restriction is $x \neq -3$ because it would make the original denominator zero.

Division by factoring is often quicker than other methods when factoring is easy or obvious. We often use polynomial division in real-world problems, too.