Lesson Video: Dividing Polynomials by Binomials Using Factorization Mathematics • 10th Grade

Video Transcript

In this video, we’ll learn how to divide polynomials by binomials using factorization. There are a number of ways to divide algebraic expressions. One of these has used a process called polynomial long division. But this can be quite a slow method. And so we should always check to see whether there are other ways that we can perform such division. One of these methods is to use factorization. Essentially, we begin by writing our division as a fraction. And then we can write both the numerator and denominator where possible in factored form.

Once we’ve done this, we simplify as we w ould with any other numeric fraction by dividing by a common factor. Now, before we go any further, it’s worth noting that this process relies on us being confident factoring algebraic expressions. These will primarily be of order two. We will also consider polynomials of order three, which can be expressed as a product of two binomials. So please ensure that you are able to factor expressions of this kind before moving any further. We‘ll now consider a simple example.

Simplify two 𝑥 squared plus five 𝑥 minus three over 𝑥 plus three.

First, we recall that this fraction line means divide. And so when we simplify, we’re actually saying how can we divide two 𝑥 squared plus five 𝑥 minus three by 𝑥 plus three. Well, once our division problem is written as a fraction, we look to factor where possible. Now, it’s not possible to factor the expression on the denominator. But we can factor the numerator. Let’s look to factor two 𝑥 squared plus five 𝑥 minus three. There are a number of ways of doing this. One way is called the AC method. It’s called the AC method because given a quadratic equation of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, we begin by multiplying the value of 𝑎 and 𝑐. In our equation, 𝑎, which is the coefficient of 𝑥 squared, is two and 𝑐 is negative three. Two multiplied by negative three is negative six.

Our next step, is just like when we factor a quadratic equation where the coefficient of 𝑥 squared is one. We look for two numbers that multiply to make negative six and add to make five. Well, six multiplied by negative one is negative six. And six plus negative one is five. And so we’re going to look to split this middle term up into six 𝑥 and negative one 𝑥. We now write our quadratic as two 𝑥 squared plus six 𝑥 minus one equals three. Now, we’ve not done anything mind-blowing here. We’ve just rewritten our original expression. If we were to now simplify the expression on the right-hand side, that would take us back to the expression on the left.

The next step is to consider the two pairs of terms. We’re going to factor each pair. We see that two 𝑥 squared and six 𝑥 have a highest common factor or a greatest common factor of two 𝑥. And so two 𝑥 squared plus six 𝑥 can be written as two 𝑥 times 𝑥 plus three. Similarly, negative one 𝑥 minus three have a common factor of negative one. So when we factor this expression, we get negative one 𝑥 plus three.

Notice now that each term contains a factor of 𝑥 plus three. And so we can factor by 𝑥 plus three. Two 𝑥 times 𝑥 plus three divided by 𝑥 plus three gives us two 𝑥. Then negative one times 𝑥 plus three divided by 𝑥 plus three gives us negative one. And so we fully factored our quadratic. It’s 𝑥 plus three times two 𝑥 minus one. And this is great because we could now rewrite our fraction. We’ve replaced the quadratic with its factored form. And we see it’s equal to 𝑥 plus three times two 𝑥 minus one all over 𝑥 plus three.

Now that it’s in this form, we can simplify our fraction as we would in numerical fraction by dividing through by any common factors. In this case, we can divide through by 𝑥 plus three. When we do, we see that our expression fully simplifies to two 𝑥 minus one over one or simply two 𝑥 minus one. And so the answer to our question and, in fact, the answer to two 𝑥 squared plus five 𝑥 minus three divided by 𝑥 plus three is two 𝑥 minus one. Now, we did use something called the AC method to factor our quadratic expression. You may be used to using some other method. And that’s absolutely fine as long as you do indeed end up with 𝑥 plus three times two 𝑥 minus one.

We’ll now look at how to find the denominator of a fraction when dividing a polynomial by a binomial using factorization.

Find the denominator of the fraction in this equation: Three 𝑥 squared plus 11𝑥 plus eight over what is equal to 𝑥 plus one.

Let’s call the denominator of our fraction 𝑦 for now, where 𝑦 is going to be some function of 𝑥. Then we’re going to recall the relationship between fractions and division. A fraction is just another way of writing a division. So what this question is also asking us is, what is the value of 𝑦 such that three 𝑥 squared plus 11𝑥 plus eight divided by 𝑦 equals 𝑥 plus one? Now, we could rearrange this to make 𝑦 the subject. Or we can quote that if 𝑎 divided by 𝑏 is equal to 𝑐, then 𝑎 divided by 𝑐 must be equal to 𝑏. This makes a lot of sense because if we rearrange each equation, we find that 𝑎 is equal to 𝑏 times 𝑐. We can think of 𝑏 and 𝑐 as a factor pair of 𝑎. We can therefore say that three 𝑥 squared plus 11𝑥 plus eight divided by 𝑥 plus one must be equal to 𝑦.

But how do we work out this division on the lef-hand side? Well, we have a number of methods, but factorization is generally the most straightforward. We’re going to look to factor the expression three 𝑥 squared plus 11𝑥 plus eight. There are a number of ways to do this. One method is kind of observation. It’s a quadratic equation, and there are no common factors apart from one in each of our terms. And so we know we can write it as the product of two binomials. The first term in each binomial must be three 𝑥 and 𝑥 because three 𝑥 times 𝑥 gives us the three 𝑥 squared we need.

And then we need to look for factor pairs of eight bearing in mind that one of these is going to be multiplied by three. And then once that happens, when we add our numbers together, we’re going to get 11. Well, a factor pair we could use is eight and one. And if we multiply three by one, we get three. Then three plus eight is 11. For this to work, both eight and one need to be positive. And so we factored our expression. It’s three 𝑥 plus eight times 𝑥 plus one. And so we can now rewrite our equation as three 𝑥 plus eight times 𝑥 plus one all over 𝑥 plus one equals 𝑦.

Now, our next step since it’s written as a fraction is to simplify like we would any other fraction by dividing through by a common factor. Here, we see we have a common factor of 𝑥 plus one. 𝑥 plus one divided by 𝑥 plus one is simply one. And so we see that 𝑦 is equal to three 𝑥 plus eight? And since we said 𝑦 was the denominator of our fraction, then the denominator is three 𝑥 plus eight. Now, a really quick way to check this answer is to check that the product of 𝑥 plus one and our denominator is indeed equal to three 𝑥 squared plus 11𝑥 plus eight. And in fact, if we multiply these two binomials, we do indeed get three 𝑥 squared plus 11𝑥 plus eight.

In our next example, we’ll look at how we can use a similar method to help us find the value of an unknown.

Find the value of 𝑘 that makes the expression 𝑥 squared minus 𝑘𝑥 plus 30 divisible by 𝑥 minus five.

When we divide polynomials by binomials, we look to begin by writing them as a fraction and then simplifying as far as possible. And so to divide 𝑥 squared minus 𝑘𝑥 plus 30 by 𝑥 minus five, we begin by writing it as 𝑥 squared minus 𝑘𝑥 plus 30 over 𝑥 minus five. And so the implication is that for our expression to be divisible by 𝑥 minus five, this fraction can be simplified. And we simplify, of course, by dividing through by a common factor. On the denominator of our fraction, we have 𝑥 minus five. So that must indicate to us that 𝑥 minus five is a factor of 𝑥 squared minus 𝑘𝑥 plus 30. So we should be able to write 𝑥 squared minus 𝑘𝑥 plus 30 as some binomial — I’ve called that 𝑥 plus 𝑎, where 𝑎 is a constant — times 𝑥 minus five.

But how do we decide what 𝑎 needs to be? We think about how we factor quadratic expressions where the coefficient of 𝑥 squared is equal to one. We have an 𝑥 at the front of each binomial. And then we look for two numbers whose product is the constant — so here, that would be 30 — and whose sum is the coefficient of 𝑥. So here, that’s negative 𝑘. These two numbers give us the numerical parts of our binomials. And so whilst we don’t know what 𝑘 is, we can say that 𝑎 multiplied by negative five must be 30. And so we’ll solve for 𝑎 by dividing through by negative five. 30 divided by negative five is negative six. So we say that 𝑎 is equal to negative six.

If we replace 𝑎 with negative six — and now forget about the denominators because they’re of course equal — we see that the numerators must be equal. We see that 𝑥 squared minus 𝑘𝑥 plus 30 must be equal to 𝑥 minus six times 𝑥 minus five. If we distribute these parentheses and simplify, that will tell us the value of 𝑘. So we begin by multiplying the first term in each binomial. 𝑥 times 𝑥 is 𝑥 squared. We then multiply the outer terms to give us negative five 𝑥, then multiply the inner terms to give us negative six 𝑥. Finally, we multiply the last terms. Negative six times negative five is 30. So we get 𝑥 squared minus five 𝑥 minus six 𝑥 plus 30. And since negative five 𝑥 minus six 𝑥 is negative 11𝑥, this becomes 𝑥 squared minus 11𝑥 plus 30.

Let’s compare both sides of this equation. We have 𝑥 squared on both sides, and we have plus 30. And so we can say that these two terms must be equal. Negative 𝑘𝑥 must be equal to negative 11𝑥. Well, for this to be true, 𝑘 must therefore be equal to 11. The value of 𝑘 that makes the expression 𝑥 squared minus 𝑘𝑥 plus 30 divisible by 𝑥 minus five is 11.

We’ll now look at how this process can help us solve problems involving geometry.

A rectangle has an area of 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10 square centimeters and a width of 𝑦 plus two centimeters. Find its length in terms of 𝑦 and its perimeter when 𝑦 equals four.

Our rectangle could look a little something like this. We know it has a width of 𝑦 plus two centimeters, and we’re trying to find its length in terms of 𝑦. For now, we’ll call its length 𝑥 centimeters, where 𝑥 is going to be some function in 𝑦. Now, we know that the area of a rectangle is given by multiplying its width by its length. So in this case, the area of the rectangle would be 𝑦 plus two times 𝑥. Let’s write that as 𝑥 times 𝑦 plus two. But we’re actually given an expression for the area. It’s 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10. And so we actually have an equation that we can look to solve or at least make 𝑥 the subject.

We’re going to make 𝑥 the subject by dividing both sides by 𝑦 plus two. On the right-hand side, that leaves us with 𝑥. But what happens on the left-hand side? Well, for now, we’ll write it as a fraction, since a fraction line simply means divide. And one of the best ways that we have to simplify a fraction, which is the same as dividing, is to begin by factoring where possible. Let’s factor the expression 𝑦 cubed plus two 𝑦 squared plus five 𝑦 plus 10. To do this, we begin by factoring the pairs of terms. When we factor 𝑦 cubed plus two 𝑦 squared, we get 𝑦 squared times 𝑦 plus two.

Similarly, when we factor five 𝑦 plus 10, we get five times 𝑦 plus two. We now know that there is a common factor of 𝑦 plus two in our two terms. And so we can factor by 𝑦 plus two. 𝑦 squared times 𝑦 plus two divided by 𝑦 plus two is 𝑦 squared. Then when we divide our second term, five times 𝑦 plus two, by 𝑦 plus two, we’re simply left with five. This means we can rewrite our fraction as 𝑦 plus two times 𝑦 squared plus five over 𝑦 plus two. And now, we see that there is a common factor of 𝑦 plus two on both the numerator and denominator of our fraction. And so we’re going to divide through by 𝑦 plus two. And when we do, we find that 𝑥 is equal to 𝑦 squared plus five.

Remember, we said that the length of our rectangle was 𝑥 centimeters. So in fact, we’ve discovered that the length in terms of 𝑦 is 𝑦 squared plus five centimeters. Now, we’re not quite finished. The question wants us to find the perimeter of this rectangle when 𝑦 is equal to four. So we’re going to substitute 𝑦 equals four into the expressions for the width and length. The width becomes four plus two, which is six centimeters. And the length becomes four squared plus five, which is 21 centimeters. The perimeter is the entire distance around the outside. So we’re going to add 21 and 6 and then multiply that by two to represent the other two sides. 21 plus six is 27, and 27 times two is 54. The perimeter of our rectangle is therefore 54 centimeters.

In this video, we saw that to divide a polynomial by a binomial, one of the most efficient methods is to use factorization. When using this method, the first thing that we do is we write our division as a fraction. The dividend, that’s the expression that we’re dividing into, is the numerator of our fraction. And the divisor, which is what we’re dividing by, is the denominator. We then factor where possible. And in this video, we’ve mainly factored the numerators. But there will be instances where we need to factor the denominator too. Once we’ve done this, we look for any common factors and we divide through. This is the same as simplifying any normal fraction. And once it’s fully simplified, we’ve performed our division.