Lesson Video: Factoring Nonmonic Quadratics Mathematics

Video Transcript

Factoring Nonmonic Quadratics

In this video, we’re going to learn a method which will help us to factor quadratic expressions in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, where the coefficient of our leading term 𝑎 is not equal to one. We’ll then see how we can use this to help us solve certain types of equations.

To start, we recall an expression in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, where 𝑎, 𝑏, and 𝑐 are constants, is called a quadratic expression. And there’s one thing worth noting here. We also assume the value of 𝑎 is not equal to zero because otherwise we would then have a linear expression. And we’ve already seen how to factor some of these expressions. For example, if the value of 𝑎 is equal to one, we know how to factor 𝑥 squared plus 𝑏𝑥 plus 𝑐. And to do this, we find two numbers which multiply to give us the constant term of 𝑐 and add to give us the coefficient of our 𝑥-term, 𝑏. 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.

We’re going to want to do the same thing to try and help us factor even more quadratic expressions. In particular, we’re going to want to apply this to quadratic expressions we have not applied them to before. We’re 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 𝑎 is not allowed to be equal to one. So we want to factor nonmonic quadratic expressions. And we’re going to do this by first starting with the factored form of our quadratic.

So let’s clear some space. And we’re going to start by writing down our factors. Remember, these need to be factors of 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. By using the FOIL method, we know if we want two factors of this expression, we’re going to need 𝑥-terms in both of these so that we get an 𝑥 squared term. And we also need a constant term. So we’re going to need to add constant terms to both of our factors. We’ll call these 𝛽 and 𝛿. In fact, we can call these variables whatever we want except the variables 𝑎, 𝑏, and 𝑐 because we’ve already used these in our quadratic.

But we’re not done yet. If we were to multiply this expression out by using the FOIL method, we would only get the term 𝑥 squared. However, our leading coefficient is 𝑎, not one. So we need to include coefficients of 𝑥. And we’ll call these 𝛼 and 𝛾. And remember, these need to be factors of our quadratic expression. So when we expand this out, we should get 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. So let’s 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 𝛼𝑥 times 𝛾𝑥, which we can simplify as 𝛼 times 𝛾 times 𝑥 squared. Next, the FOIL method tells us we want to multiply the outer two terms together. This gives us 𝛼𝑥 multiplied by 𝛿, and we can rearrange this to give us 𝛼 times 𝛿 times 𝑥. Next, the FOIL method tells us we need to multiply our inner two terms together. This of course gives us 𝛽 multiplied by 𝛾 times 𝑥. 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 𝛽 multiplied by 𝛿. 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 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. And now we can start seeing useful information. For example, if we compare the leading coefficients, we see that 𝑎 should be equal to 𝛼 times 𝛾. Similarly, comparing the constant two terms, we should see that 𝑐 should be equal to 𝛽 times 𝛿. But what information can we learn about the coefficient of 𝑥? In our expansion, there’s two terms of 𝑥. So we’re going to need to simplify this.

We’re going to want to take out the shared factor of 𝑥. Doing this, we get 𝛼𝛿 plus 𝛽𝛾 all multiplied by 𝑥. And now we can compare the coefficients of 𝑥. We see we split 𝑏 into two numbers which add together to give us 𝑏. 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 we’re doing is changing the value of 𝑏 to be the sum of two values. So to accomplish our first step, we need to find the values of 𝛼 times 𝛿 and 𝛽 times 𝛾. So let’s try and find these values. Well, we already know when we add them together, we should get the coefficient of 𝑥.

But of course this is not enough information to find these values because there’s a lot of different ways of adding two numbers together to give us 𝑏. So in fact, we need more information to find these values. And to see this, let’s see what happens if we were to multiply these two values together. So let’s multiply these two terms together, 𝛼 times 𝛿 multiplied by 𝛽 times 𝛾. At first, this doesn’t seem very useful. However, we can notice something very interesting about this expression. It contains 𝛼 multiplied by 𝛾. And we know that 𝛼 multiplied by 𝛾 is the coefficient of 𝑥 squared. It’s just equal to 𝑎. Similarly, we’ve already found an expression for 𝛽 multiplied by 𝛿. It’s the value 𝑐. So in actual fact, the product of these two numbers is just equal to 𝑎 times 𝑐. And this is enough to help us start factoring our quadratic.

We need to find two numbers that add to give us 𝑏 and multiply to give us 𝑎 times 𝑐. And the easiest way to do this is to find the factor pairs of 𝑎 times 𝑐. 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 𝑎 times 𝑐. Then, in this list of factors, we need to find the pair which add to give us our coefficient of 𝑥. This is equivalent to finding the values of 𝛼 times 𝛿 and 𝛽 times 𝛾. We can then use these to split our coefficient of 𝑥. So we want to write 𝑏 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 won’t be able to directly go from one step to the other. Instead, it’s easier to think of this as two separate expressions, where we’re going to want to take out the constant factors. So we’re going to want to factor 𝑎𝑥 squared plus 𝛼𝛿𝑥 and 𝛽𝛾𝑥 plus 𝛽𝛿 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 we’ll have successfully fully factored our nonmonic quadratic. This is a very complicated process to explain. However, it’s a lot easier to see this used in an example.

Factor six 𝑥 squared minus five 𝑥 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; it’s not equal to one. So this is a nonmonic quadratic. So to fully factor this, we’re going to need to recall how we fully factor nonmonic quadratic expressions. The first thing we do whenever we’re asked to factor a quadratic 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is find all of the factor pairs of 𝑎 times 𝑐. In our case, our value of 𝑎, the leading coefficient, is six, and our value of 𝑐, the constant term, is negative four. So 𝑎 times 𝑐 is going to be six multiplied by negative four, and we can evaluate this. It’s 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, it’s 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 we’re 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 𝑥, which is 𝑏. And we can see in the quadratic given to us in the question, the coefficient of 𝑥 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 there’s only one option: three add negative eight is equal to negative five. So we’re going to want to use the values of three and negative eight. And although it’s not necessary, it can be useful to give these letters. We’ll set 𝑛 equal to three and 𝑚 equal to negative eight.

The third step in our factorization process will be splitting our value of 𝑏 by using our values of 𝑛 and 𝑚. We want to rewrite our quadratic as 𝑎𝑥 squared plus 𝑛𝑥 plus 𝑚𝑥 plus 𝑐. In our case, the value of 𝑎 is six, the value of 𝑏 is negative five, and the value of 𝑐 is negative four. And now all we’re 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 𝑥 squared plus three 𝑥 minus eight 𝑥 minus four.

Our fourth step tells us to factor 𝑎𝑥 squared plus 𝑛𝑥 and 𝑚𝑥 plus 𝑐 separately. And to do this, it can be sometimes helpful to think of splitting our expression in half. And it’s also important to note we want to do this so that our two-halves have a common factor. So let’s try and factor each of these halves separately. First, in our first half, we can see that six 𝑥 squared and three 𝑥 share a factor of three 𝑥. So let’s take out the shared factor of three 𝑥 in our first two terms. We have three 𝑥 multiplied by two 𝑥 gives us our first term of six 𝑥 squared. And three 𝑥 multiplied by one gives us our second term of three 𝑥. So we factor the first two terms to give us three 𝑥 times two 𝑥 plus one.

We now want this same expression of two 𝑥 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 we’re going to take out the shared factor of negative four in our third and fourth term. We have negative four multiplied by two 𝑥 gives us negative eight 𝑥, and negative four multiplied by one gives us negative four. Therefore, we’ve rewritten our quadratic as three 𝑥 times two 𝑥 plus one minus four times two 𝑥 plus one.

And now hopefully, we can see the shared factor of two 𝑥 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 we’re taking out our factor of two 𝑥 plus one. We need to multiply our first term by three 𝑥 and our second term by negative four. This gives us two 𝑥 plus one multiplied by three 𝑥 minus four. And this gives us our final answer. We were able to show six 𝑥 squared minus five 𝑥 minus four is equal to two 𝑥 plus one times three 𝑥 minus four.

And there’s a few things worth pointing out here. Just like with monic quadratics, we can’t 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, there’s one interesting question. What would’ve happened if we had changed the order of 𝑛 and 𝑚? If we had changed our values of 𝑛 and 𝑚 to be the other way round, when we split our coefficient of 𝑏, we would instead get negative eight 𝑥 plus three 𝑥. 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 𝑥. Taking out this factor of two 𝑥, we would get two 𝑥 times three 𝑥 minus four. We then want this factor of three 𝑥 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 we’ll write the third and fourth term as one times three 𝑥 minus four. Now, we can see that three 𝑥 minus four is a shared factor. And if we take out this factor of three 𝑥 minus four, we get three 𝑥 minus four multiplied by two 𝑥 plus one, which is the exact same answer we had before. However, our factors are in the opposite order. In fact, it doesn’t matter which one we call 𝑛 and which one we call 𝑚. We’ll always get the same answer.

Therefore, we were able to fully factor six 𝑥 squared minus five 𝑥 minus four. We got two 𝑥 plus one times three 𝑥 minus four.

Let’s now see another example of factoring a nonmonic quadratic.

Factorize fully negative 45 minus 24𝑥 plus 48𝑥 squared.

We’re given a quadratic equation which we’re 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, we’ll also take this opportunity to rewrite our quadratic in decreasing powers of 𝑥. So let’s start by taking three out of 48𝑥 squared. Of course, three times 16𝑥 squared is equal to 48𝑥 squared. Next, three multiplied by negative eight 𝑥 is equal to negative 24𝑥. Finally, three multiplied by negative 15 is equal to negative 45. So now instead we can factor 16𝑥 squared minus eight 𝑥 minus 15.

To factor our quadratic 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, the first step is to find the factor pairs of 𝑎 times 𝑐. In this case, the value of 𝑎 is 16 and the value of 𝑐 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 we’ll move onto the second step.

The second step is to choose a factor pair whose sum is equal to 𝑏. And in our case, the value of 𝑏 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 𝑛 is 12 and 𝑚 is negative 20.

Step three is to split our value of 𝑏 into 𝑛 plus 𝑚. By doing this, we can rewrite our quadratic as 16𝑥 squared plus 12𝑥 minus 20𝑥 minus 15. And remember, our original quadratic was this multiplied by three. So we’ll 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 16𝑥 squared and 12𝑥 share a factor of four 𝑥. So we can factor this to get four 𝑥 times four 𝑥 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 𝑥 plus three. And don’t 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 𝑥 plus three multiplied by four 𝑥 minus five.

Let’s now see how we could use this to solve an equation. We can ask the question “Solve the equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero,” where of course our value of 𝑎 is not equal to zero. This means we want to find all of the values of 𝑥 such that when we substitute them into our quadratic, it’s equal to zero.

Now we can’t find these values of 𝑥 directly from our quadratic. Instead, we’re going to need to try factoring our quadratic. We’ve seen, under certain conditions, we can fully factor our quadratic. We can factor our quadratic to get 𝛼𝑥 plus 𝛽 times 𝛾𝑥 plus 𝛿 for some constants 𝛼, 𝛽, 𝛾, and 𝛿. And this is in fact true even if our value of 𝑎 is equal to one. If this is the case, then we can set 𝛼 and 𝛾 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 let’s 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 𝛼𝑥 plus 𝛽 is equal to zero or 𝛾𝑥 plus 𝛿 is equal to zero. And we can just rearrange these equations to find our values of 𝑥. Setting our first factor equal to zero, subtracting 𝛽 from both sides of the equation, and then dividing through by 𝛼, we get 𝑥 is equal to negative 𝛽 divided by 𝛼.

And it’s also worth pointing out here we know that both 𝛼 and 𝛾 are not equal to zero because, remember, 𝛼 multiplied by 𝛾 must be equal to our coefficient of 𝑥 squared. It must be equal to 𝑎. This means we’re allowed to divide through by 𝛼 and 𝛾.

We can now do the same for our second factor. Setting our second factor equal to zero, subtracting 𝛿 from both sides, and dividing through by 𝛾, we get 𝑥 is equal to negative 𝛿 over 𝛾. So we’ve 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.

Let’s now go through the key points we saw throughout this video. First, we know an expression in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, where 𝑎 is not equal to zero, is called a quadratic. And we can call it a nonmonic quadratic if the coefficient of the leading term 𝑎 is not equal to one. And we’ll sometimes call quadratics monic if 𝑎 is equal to one. Next, it’s important to note that we can’t always factor quadratics into linear factors. A good example of this is the quadratic 𝑥 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 𝑎 multiplied by 𝑐. Next, we saw we need to find the pair which adds to give us our coefficient 𝑏. And we also saw it can be useful to call these 𝑛 and 𝑚. And we also saw it doesn’t matter which of our pair we call 𝑛 and which of our pair we call 𝑚. It won’t change our answer at the end. Next, we saw to factor our quadratic we want to split our 𝑥-term by using that 𝑏 is equal to 𝑛 plus 𝑚. So 𝑏𝑥 will be equal to 𝑛𝑥 plus 𝑚𝑥. The fourth thing we’re going to need to do is factor 𝑎𝑥 squared plus 𝑛𝑥 and 𝑚𝑥 plus 𝑐 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 it’s 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 𝛼𝑥 plus 𝛽 times 𝛾𝑥 plus 𝛿, 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 𝑥 is equal to negative 𝛽 over 𝛼 or 𝑥 is equal to negative 𝛿 over 𝛾.