### Video Transcript

In this video, weβll 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 weβve 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, itβs
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. Weβll now consider a simple
example.

Simplify two π₯ squared plus five
π₯ minus three over π₯ plus three.

First, we recall that this fraction
line means divide. And so when we simplify, weβre
actually saying how can we divide two π₯ squared plus five π₯ minus three by π₯ plus
three. Well, once our division problem is
written as a fraction, we look to factor where possible. Now, itβs not possible to factor
the expression on the denominator. But we can factor the
numerator. Letβs look to factor two π₯ squared
plus five π₯ minus three. There are a number of ways of doing
this. One way is called the AC
method. Itβs called the AC method because
given a quadratic equation of the form ππ₯ squared plus ππ₯ plus π, we begin by
multiplying the value of π and π. In our equation, π, which is the
coefficient of π₯ squared, is two and π 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 π₯ 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 weβre going to look to split
this middle term up into six π₯ and negative one π₯. We now write our quadratic as two
π₯ squared plus six π₯ minus one equals three. Now, weβve not done anything
mind-blowing here. Weβve 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. Weβre going to factor each
pair. We see that two π₯ squared and six
π₯ have a highest common factor or a greatest common factor of two π₯. And so two π₯ squared plus six π₯
can be written as two π₯ times π₯ plus three. Similarly, negative one π₯ minus
three have a common factor of negative one. So when we factor this expression,
we get negative one π₯ plus three.

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

Now that itβs 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 π₯ plus three. When we do, we see that our
expression fully simplifies to two π₯ minus one over one or simply two π₯ minus
one. And so the answer to our question
and, in fact, the answer to two π₯ squared plus five π₯ minus three divided by π₯
plus three is two π₯ 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 thatβs absolutely fine as long
as you do indeed end up with π₯ plus three times two π₯ minus one.

Weβll 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 π₯ squared plus 11π₯ plus eight over what is equal
to π₯ plus one.

Letβs call the denominator of our
fraction π¦ for now, where π¦ is going to be some function of π₯. Then weβre 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 π¦ such that three π₯ squared plus 11π₯ plus
eight divided by π¦ equals π₯ plus one? Now, we could rearrange this to
make π¦ the subject. Or we can quote that if π divided
by π is equal to π, then π divided by π must be equal to π. This makes a lot of sense because
if we rearrange each equation, we find that π is equal to π times π. We can think of π and π as a
factor pair of π. We can therefore say that three π₯
squared plus 11π₯ plus eight divided by π₯ plus one must be equal to π¦.

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. Weβre going to look to factor the
expression three π₯ squared plus 11π₯ plus eight. There are a number of ways to do
this. One method is kind of
observation. Itβs 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 π₯ and π₯ because three π₯ times π₯ gives us the three π₯ 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, weβre 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. Itβs three π₯ plus eight times π₯
plus one. And so we can now rewrite our
equation as three π₯ plus eight times π₯ plus one all over π₯ plus one equals
π¦.

Now, our next step since itβs
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 π₯ plus one. π₯ plus one divided by π₯ plus one
is simply one. And so we see that π¦ is equal to
three π₯ plus eight? And since we said π¦ was the
denominator of our fraction, then the denominator is three π₯ plus eight. Now, a really quick way to check
this answer is to check that the product of π₯ plus one and our denominator is
indeed equal to three π₯ squared plus 11π₯ plus eight. And in fact, if we multiply these
two binomials, we do indeed get three π₯ squared plus 11π₯ plus eight.

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

Find the value of π that makes the
expression π₯ squared minus ππ₯ plus 30 divisible by π₯ 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 π₯ squared minus
ππ₯ plus 30 by π₯ minus five, we begin by writing it as π₯ squared minus ππ₯ plus
30 over π₯ minus five. And so the implication is that for
our expression to be divisible by π₯ 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 π₯ minus five. So that must indicate to us that π₯
minus five is a factor of π₯ squared minus ππ₯ plus 30. So we should be able to write π₯
squared minus ππ₯ plus 30 as some binomial β Iβve called that π₯ plus π, where π
is a constant β times π₯ minus five.

But how do we decide what π needs
to be? We think about how we factor
quadratic expressions where the coefficient of π₯ squared is equal to one. We have an π₯ at the front of each
binomial. And then we look for two numbers
whose product is the constant β so here, that would be 30 β and whose sum is the
coefficient of π₯. So here, thatβs negative π. These two numbers give us the
numerical parts of our binomials. And so whilst we donβt know what π
is, we can say that π multiplied by negative five must be 30. And so weβll solve for π by
dividing through by negative five. 30 divided by negative five is
negative six. So we say that π is equal to
negative six.

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

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

Weβll now look at how this process
can help us solve problems involving geometry.

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

Our rectangle could look a little
something like this. We know it has a width of π¦ plus
two centimeters, and weβre trying to find its length in terms of π¦. For now, weβll call its length π₯
centimeters, where π₯ is going to be some function in π¦. 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 π¦ plus two times π₯. Letβs write that as π₯ times π¦
plus two. But weβre actually given an
expression for the area. Itβs π¦ cubed plus two π¦ squared
plus five π¦ plus 10. And so we actually have an equation
that we can look to solve or at least make π₯ the subject.

Weβre going to make π₯ the subject
by dividing both sides by π¦ plus two. On the right-hand side, that leaves
us with π₯. But what happens on the left-hand
side? Well, for now, weβll 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. Letβs factor the expression π¦
cubed plus two π¦ squared plus five π¦ plus 10. To do this, we begin by factoring
the pairs of terms. When we factor π¦ cubed plus two π¦
squared, we get π¦ squared times π¦ plus two.

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

Remember, we said that the length
of our rectangle was π₯ centimeters. So in fact, weβve discovered that
the length in terms of π¦ is π¦ squared plus five centimeters. Now, weβre not quite finished. The question wants us to find the
perimeter of this rectangle when π¦ is equal to four. So weβre going to substitute π¦
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 weβre 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, thatβs the expression
that weβre dividing into, is the numerator of our fraction. And the divisor, which is what
weβre dividing by, is the denominator. We then factor where possible. And in this video, weβve mainly
factored the numerators. But there will be instances where
we need to factor the denominator too. Once weβve done this, we look for
any common factors and we divide through. This is the same as simplifying any
normal fraction. And once itβs fully simplified,
weβve performed our division.