Video Transcript
Factoring Nonmonic Quadratics
In this video, weβre going to learn
a method which will help us to factor quadratic expressions in the form ππ₯ squared
plus ππ₯ plus π, where the coefficient of our leading term π is not equal to
one. Weβll then see how we can use this
to help us solve certain types of equations.
To start, we recall an expression
in the form ππ₯ squared plus ππ₯ plus π, where π, π, and π are constants, is
called a quadratic expression. And thereβs one thing worth noting
here. We also assume the value of π is
not equal to zero because otherwise we would then have a linear expression. And weβve already seen how to
factor some of these expressions. For example, if the value of π is
equal to one, we know how to factor π₯ squared plus ππ₯ plus π. And to do this, we find two numbers
which multiply to give us the constant term of π and add to give us the coefficient
of our π₯-term, π. And in fact we managed to prove
this directly. We did this by instead of starting
with our quadratic we instead started with the factored form and worked our way
backwards. This then gave us information about
our factors.
Weβre going to want to do the same
thing to try and help us factor even more quadratic expressions. In particular, weβre going to want
to apply this to quadratic expressions we have not applied them to before. Weβre going to call these nonmonic
quadratic expressions. The word monic means one. When we say a nonmonic quadratic
expression, we mean a quadratic expression where its leading coefficient is not
equal to one. In other words, our value of π is
not allowed to be equal to one. So we want to factor nonmonic
quadratic expressions. And weβre going to do this by first
starting with the factored form of our quadratic.
So letβs clear some space. And weβre going to start by writing
down our factors. Remember, these need to be factors
of ππ₯ squared plus ππ₯ plus π. By using the FOIL method, we know
if we want two factors of this expression, weβre going to need π₯-terms in both of
these so that we get an π₯ squared term. And we also need a constant
term. So weβre going to need to add
constant terms to both of our factors. Weβll call these π½ and πΏ. In fact, we can call these
variables whatever we want except the variables π, π, and π because weβve already
used these in our quadratic.
But weβre not done yet. If we were to multiply this
expression out by using the FOIL method, we would only get the term π₯ squared. However, our leading coefficient is
π, not one. So we need to include coefficients
of π₯. And weβll call these πΌ and πΎ. And remember, these need to be
factors of our quadratic expression. So when we expand this out, we
should get ππ₯ squared plus ππ₯ plus π. So letβs multiply these two
binomials together by using the FOIL method.
The FOIL method tells us we need to
start by multiplying the first two terms of our binomials together. This gives us πΌπ₯ times πΎπ₯,
which we can simplify as πΌ times πΎ times π₯ squared. Next, the FOIL method tells us we
want to multiply the outer two terms together. This gives us πΌπ₯ multiplied by
πΏ, and we can rearrange this to give us πΌ times πΏ times π₯. Next, the FOIL method tells us we
need to multiply our inner two terms together. This of course gives us π½
multiplied by πΎ times π₯. Finally, we need to add on the
product of the last two terms of our binomials. And the product of the last two
terms in our factors is π½ multiplied by πΏ. So we multiplied our two factors
out to give us the following expression. And at first, it might be hard to
see how this has helped us.
But remember, we assumed this was a
factoring of our original nonmonic quadratic. So we can in fact say all of this
is equal to our original quadratic. In other words, this should be
equal to ππ₯ squared plus ππ₯ plus π. And now we can start seeing useful
information. For example, if we compare the
leading coefficients, we see that π should be equal to πΌ times πΎ. Similarly, comparing the constant
two terms, we should see that π should be equal to π½ times πΏ. But what information can we learn
about the coefficient of π₯? In our expansion, thereβs two terms
of π₯. So weβre going to need to simplify
this.
Weβre going to want to take out the
shared factor of π₯. Doing this, we get πΌπΏ plus π½πΎ
all multiplied by π₯. And now we can compare the
coefficients of π₯. We see we split π into two numbers
which add together to give us π. We want to try and reverse this
process. We want to start with our quadratic
and end up with the fully factored form. To do this, we can see in the first
step all weβre doing is changing the value of π to be the sum of two values. So to accomplish our first step, we
need to find the values of πΌ times πΏ and π½ times πΎ. So letβs try and find these
values. Well, we already know when we add
them together, we should get the coefficient of π₯.
But of course this is not enough
information to find these values because thereβs a lot of different ways of adding
two numbers together to give us π. So in fact, we need more
information to find these values. And to see this, letβs see what
happens if we were to multiply these two values together. So letβs multiply these two terms
together, πΌ times πΏ multiplied by π½ times πΎ. At first, this doesnβt seem very
useful. However, we can notice something
very interesting about this expression. It contains πΌ multiplied by
πΎ. And we know that πΌ multiplied by
πΎ is the coefficient of π₯ squared. Itβs just equal to π. Similarly, weβve already found an
expression for π½ multiplied by πΏ. Itβs the value π. So in actual fact, the product of
these two numbers is just equal to π times π. And this is enough to help us start
factoring our quadratic.
We need to find two numbers that
add to give us π and multiply to give us π times π. And the easiest way to do this is
to find the factor pairs of π times π. So in fact, we can write this
method down in steps. The first thing we want to do is
find all the pairs of numbers which multiply to give us π times π. Then, in this list of factors, we
need to find the pair which add to give us our coefficient of π₯. This is equivalent to finding the
values of πΌ times πΏ and π½ times πΎ. We can then use these to split our
coefficient of π₯. So we want to write π as the sum
of our two factors. In fact, we can just directly split
this into two terms because this is the form we want it in.
And at this point, we wonβt be able
to directly go from one step to the other. Instead, itβs easier to think of
this as two separate expressions, where weβre going to want to take out the constant
factors. So weβre going to want to factor
ππ₯ squared plus πΌπΏπ₯ and π½πΎπ₯ plus π½πΏ separately. And when we do this, both sides
will get a common factor. So the fifth and final step will
then just be to take out this common factor. And weβll have successfully fully
factored our nonmonic quadratic. This is a very complicated process
to explain. However, itβs a lot easier to see
this used in an example.
Factor six π₯ squared minus five π₯
minus four.
We can see the question is asking
us to factor an expression. And we can see that this is a
quadratic expression. However, the leading coefficient in
this expression is six; itβs not equal to one. So this is a nonmonic
quadratic. So to fully factor this, weβre
going to need to recall how we fully factor nonmonic quadratic expressions. The first thing we do whenever
weβre asked to factor a quadratic ππ₯ squared plus ππ₯ plus π is find all of the
factor pairs of π times π. In our case, our value of π, the
leading coefficient, is six, and our value of π, the constant term, is negative
four. So π times π is going to be six
multiplied by negative four, and we can evaluate this. Itβs equal to negative 24.
We want to find all of the numbers
we can multiply together to give us negative 24. And the easiest way to do this is
to start at one and move our way upwards. We know one multiplied by negative
24 is going to give us negative 24. Next, we can also pick the number
two. Two multiplied by negative 12 is
equal to negative 24. We can also try three, and we see
that three works. Three multiplied by negative eight
is equal to negative 24. And we can keep going. Four multiplied by negative six is
equal to negative 24. If we were to try five, we would
see that five does not go directly into negative 24. And then if we try six, we see that
six does.
However, we can also notice that
six already appears in our factor list. In fact, itβs negative six which
appears. And in fact this helps us a
lot. It tells us that the only other
numbers which are going to divide are eight, 12, and 24. So in fact we can write these
down. And we already know what the other
number in their factor pair is going to be. Since four times negative six is
equal to negative 24, six times negative four is going to be equal to negative
24. In fact, all weβre doing is
changing which number has the negative sign. And we can do this for the other
three numbers. We switch which number has the
negative sign. We get eight times negative three
is negative 24. 12 times negative two is equal to
negative 24. And 24 multiplied by negative one
is negative 24. So we found all of our factor
pairs.
The next step we need to do is find
the factor pair which adds to give us the coefficient of π₯, which is π. And we can see in the quadratic
given to us in the question, the coefficient of π₯ is negative five. So we need to see which of our two
factor pairs add together to give us negative five. And we can see that thereβs only
one option: three add negative eight is equal to negative five. So weβre going to want to use the
values of three and negative eight. And although itβs not necessary, it
can be useful to give these letters. Weβll set π equal to three and π
equal to negative eight.
The third step in our factorization
process will be splitting our value of π by using our values of π and π. We want to rewrite our quadratic as
ππ₯ squared plus ππ₯ plus ππ₯ plus π. In our case, the value of π is
six, the value of π is negative five, and the value of π is negative four. And now all weβre doing is using
the fact that we know negative five is equal to three plus negative eight to rewrite
our quadratic. This means our quadratic is equal
to six π₯ squared plus three π₯ minus eight π₯ minus four.
Our fourth step tells us to factor
ππ₯ squared plus ππ₯ and ππ₯ plus π separately. And to do this, it can be sometimes
helpful to think of splitting our expression in half. And itβs also important to note we
want to do this so that our two-halves have a common factor. So letβs try and factor each of
these halves separately. First, in our first half, we can
see that six π₯ squared and three π₯ share a factor of three π₯. So letβs take out the shared factor
of three π₯ in our first two terms. We have three π₯ multiplied by two
π₯ gives us our first term of six π₯ squared. And three π₯ multiplied by one
gives us our second term of three π₯. So we factor the first two terms to
give us three π₯ times two π₯ plus one.
We now want this same expression of
two π₯ plus one to appear in the right-hand side of our expression. And to do this, we need to notice
both terms share a factor of negative four. So weβre going to take out the
shared factor of negative four in our third and fourth term. We have negative four multiplied by
two π₯ gives us negative eight π₯, and negative four multiplied by one gives us
negative four. Therefore, weβve rewritten our
quadratic as three π₯ times two π₯ plus one minus four times two π₯ plus one.
And now hopefully, we can see the
shared factor of two π₯ plus one in both of these two terms. And this gives us our fifth and
final step. We need to take out this common
factor. So weβre taking out our factor of
two π₯ plus one. We need to multiply our first term
by three π₯ and our second term by negative four. This gives us two π₯ plus one
multiplied by three π₯ minus four. And this gives us our final
answer. We were able to show six π₯ squared
minus five π₯ minus four is equal to two π₯ plus one times three π₯ minus four.
And thereβs a few things worth
pointing out here. Just like with monic quadratics, we
canβt always factor nonmonic quadratics. Next, we can always check our
answer by using the FOIL method. We expand our parentheses and see
if we get the same quadratic. Finally, thereβs one interesting
question. What wouldβve happened if we had
changed the order of π and π? If we had changed our values of π
and π to be the other way round, when we split our coefficient of π, we would
instead get negative eight π₯ plus three π₯. This would then change what we need
to do in our third step.
Remember, we want to factor each
half of this expression separately. In the first two terms, we can see
they share a factor of two π₯. Taking out this factor of two π₯,
we would get two π₯ times three π₯ minus four. We then want this factor of three
π₯ minus four to appear on the right-hand side. And we can actually see this is
already exactly what we have. And one way of thinking about this
is to take out a factor of one. So weβll write the third and fourth
term as one times three π₯ minus four. Now, we can see that three π₯ minus
four is a shared factor. And if we take out this factor of
three π₯ minus four, we get three π₯ minus four multiplied by two π₯ plus one, which
is the exact same answer we had before. However, our factors are in the
opposite order. In fact, it doesnβt matter which
one we call π and which one we call π. Weβll always get the same
answer.
Therefore, we were able to fully
factor six π₯ squared minus five π₯ minus four. We got two π₯ plus one times three
π₯ minus four.
Letβs now see another example of
factoring a nonmonic quadratic.
Factorize fully negative 45 minus
24π₯ plus 48π₯ squared.
Weβre given a quadratic equation
which weβre asked to factor fully. And in fact, we can see something
interesting. We can see the leading coefficient
is not equal to one. So this is a nonmonic quadratic,
and we need to use what we know about factoring nonmonic quadratics. And we might be tempted to start
using our steps directly.
However, we can actually notice
something interesting about this quadratic. All three of the terms already
share a factor of three. So we can take out the shared
factor of three. And in fact, weβll also take this
opportunity to rewrite our quadratic in decreasing powers of π₯. So letβs start by taking three out
of 48π₯ squared. Of course, three times 16π₯ squared
is equal to 48π₯ squared. Next, three multiplied by negative
eight π₯ is equal to negative 24π₯. Finally, three multiplied by
negative 15 is equal to negative 45. So now instead we can factor 16π₯
squared minus eight π₯ minus 15.
To factor our quadratic ππ₯
squared plus ππ₯ plus π, the first step is to find the factor pairs of π times
π. In this case, the value of π is 16
and the value of π is negative 15. And this immediately runs us into a
problem. If we multiply these together, we
get negative 240. There are far too many factor pairs
here to list out. So instead weβll move onto the
second step.
The second step is to choose a
factor pair whose sum is equal to π. And in our case, the value of π is
negative eight. Now this is a very small
number. So we know that our two factors are
going to be very similar in size. If we try a few different numbers,
we can eventually find the factor pair 12 and negative 20. And we can see when we add these
two together, we get negative eight. And also the π is 12 and π is
negative 20.
Step three is to split our value of
π into π plus π. By doing this, we can rewrite our
quadratic as 16π₯ squared plus 12π₯ minus 20π₯ minus 15. And remember, our original
quadratic was this multiplied by three. So weβll just multiply both sides
of the equation through by three. Next, remember we need to factor
each side of this expression separately. First, we can see 16π₯ squared and
12π₯ share a factor of four π₯. So we can factor this to get four
π₯ times four π₯ plus three. We then want to factor the second
two terms to get this exact same common factor. We need to take out negative
five. Taking out negative five, we get
negative five times four π₯ plus three. And donβt forget we still need to
multiply this entire expression through by three. And the last thing we need to do is
take out our common factor.
And by doing this, we get our final
answer, three times four π₯ plus three multiplied by four π₯ minus five.
Letβs now see how we could use this
to solve an equation. We can ask the question βSolve the
equation ππ₯ squared plus ππ₯ plus π is equal to zero,β where of course our value
of π is not equal to zero. This means we want to find all of
the values of π₯ such that when we substitute them into our quadratic, itβs equal to
zero.
Now we canβt find these values of
π₯ directly from our quadratic. Instead, weβre going to need to try
factoring our quadratic. Weβve seen, under certain
conditions, we can fully factor our quadratic. We can factor our quadratic to get
πΌπ₯ plus π½ times πΎπ₯ plus πΏ for some constants πΌ, π½, πΎ, and πΏ. And this is in fact true even if
our value of π is equal to one. If this is the case, then we can
set πΌ and πΎ equal to one. So if we can factor our quadratic,
instead of solving the quadratic equation, we can instead solve the equation with
the factors.
But letβs see what we have
here. We have two numbers which when we
multiply them together, we get zero. And if the product of two numbers
is equal to zero, then one of our two numbers must be equal to zero. So one of our two factors must be
equal to zero. Either πΌπ₯ plus π½ is equal to
zero or πΎπ₯ plus πΏ is equal to zero. And we can just rearrange these
equations to find our values of π₯. Setting our first factor equal to
zero, subtracting π½ from both sides of the equation, and then dividing through by
πΌ, we get π₯ is equal to negative π½ divided by πΌ.
And itβs also worth pointing out
here we know that both πΌ and πΎ are not equal to zero because, remember, πΌ
multiplied by πΎ must be equal to our coefficient of π₯ squared. It must be equal to π. This means weβre allowed to divide
through by πΌ and πΎ.
We can now do the same for our
second factor. Setting our second factor equal to
zero, subtracting πΏ from both sides, and dividing through by πΎ, we get π₯ is equal
to negative πΏ over πΎ. So weβve actually found a way of
solving quadratic equations. If we can fully factor our
quadratic, we can instead solve the linear factors equal to zero.
Letβs now go through the key points
we saw throughout this video. First, we know an expression in the
form ππ₯ squared plus ππ₯ plus π, where π is not equal to zero, is called a
quadratic. And we can call it a nonmonic
quadratic if the coefficient of the leading term π is not equal to one. And weβll sometimes call quadratics
monic if π is equal to one. Next, itβs important to note that
we canβt always factor quadratics into linear factors. A good example of this is the
quadratic π₯ squared plus one. We can never write this as the
product of two linear factors.
We also found a technique to help
us factor quadratic equations. First, we need to list all of the
factor pairs of π multiplied by π. Next, we saw we need to find the
pair which adds to give us our coefficient π. And we also saw it can be useful to
call these π and π. And we also saw it doesnβt matter
which of our pair we call π and which of our pair we call π. It wonβt change our answer at the
end. Next, we saw to factor our
quadratic we want to split our π₯-term by using that π is equal to π plus π. So ππ₯ will be equal to ππ₯ plus
ππ₯. The fourth thing weβre going to
need to do is factor ππ₯ squared plus ππ₯ and ππ₯ plus π separately to find a
common factor.
And we also saw it can be useful
sometimes to think of this as splitting our expression in half. And itβs very important that we
factor this in such a way that both sides share a common factor. And then finally, we need to take
out the common factor from both sides of this expression. Then, our quadratic will be fully
factored.
And the last thing we saw is that
we can solve a quadratic equation is equal to zero by fully factoring our
quadratic. If we can factor our quadratic into
the form πΌπ₯ plus π½ times πΎπ₯ plus πΏ, then the product of these is equal to
zero. And for the product of two numbers
to be equal to zero, one of our numbers must be equal to zero. So we can solve this by solving our
linear factors are equal to zero. We get π₯ is equal to negative π½
over πΌ or π₯ is equal to negative πΏ over πΎ.