 Welcome back to our lecture series math 1050 college algebra for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In our penultimate section for chapter 7 about pre-calculus, I want to review in this video some different techniques we can use to solve equations. And in particular, I want to talk about solving equations by factoring. There are lots of equations for which you can just perform inverse operations over and over and over again until you solve them. So, like, if you were given something like y equals 3 times the natural log of x minus 1 plus 4 equals 0, let's just say that so that we don't have the variable y in play. We have an equation like this, right? You can just start peeling off operations one by one by one, right? You're going to subtract 4 from both sides. You're going to divide both sides by 3. You're going to exponentiate both sides by the power of e and then you're going to add 1 to both sides. I can tell you exactly what happens and then x would be the solution in the end like that. So, many equations, you can just perform inverse operations over and over and over again to solve it, okay? Another thing I want to mention here is when it comes to solving equations, I'm going to emphasize the case where the right-hand side equals 0. Because if that wasn't the case, like, if you had, like, some function, you know, some algebraic expression on the left-hand side and some algebraic expression on the right-hand side, you could always put it equal to 0, right, just by subtracting things over. And so, if you can solve an equation for which the right-hand side equals 0, you can solve any equation. So, we're going to use that special assumption that the right-hand side equals 0 in this consideration. Now, some other things we saw as we solve equations, we might have to clear denominators, right? We might have to, because the thing is what happens is our variable in play here sometimes is in prison in things, right? Like, you know, if your x is inside of some parentheses, you might have to distribute to get it out. Or if there's, like, a power, you might have to foil it to get it out. Or if it's in a denominator, you might have to clear the denominators. There are some strategies to liberate your variable x from some type of prism, prison of some kind. And I'm not going to talk so much about that in this video. We've done things like that before. A few examples might come up here. But really, in this video, I want to emphasize how useful factoring can be in a tool, as a tool to solve equations. In particular, the reason why this is so useful, I should say, is in the calculus setting, oftentimes you have to set your so-called derivative equal to 0. Don't worry about what a derivative is at this moment, but it's something you care about a lot in calculus. You have to set your derivative equal to 0. And because of the way you computed the derivative, the derivative often has, that is, we can solve these equations often by factoring. So if I just give you a random equation, factoring might seem kind of limited in its use. But when you're trying to find critical numbers of a derivative, it turns out factoring shows up a lot. It turns out to be very useful by the nature of how one computes the derivative. So why is factoring a useful tool for solving equations? Well, it comes down to the zero product property we've seen before. That if you have real numbers A and B, the only way the product A, B equals 0 is if A were equal to 0 or B equals 0. I guess they could both be 0, right? But one of the factors had to equal 0. So this property allows us to use factoring to solve many, many types of equations of the form f of x equals 0. Literally, like we said earlier, setting the right hand side equal to 0 is actually not a huge assumption whatsoever. We can always make it that way. And in fact, in calculus, like I said, on one of these problems that you probably will see, turns out factoring is very useful in solving this equation f of x equals 0. It's a very useful tool. So notice graphically solving f of x equals 0 is just finding the x intercepts of the graph. It's equally important also to determine the domain of f to know exactly where the expression f of x is undefined. We often write this as does not exist, D and E, right? These are so-called discontinuities, right? We've seen pictures of graphs that have like a vertical asymptote. Maybe there's some type of like jump discontinuity that happened, right? These places are concerning to us as well. So when it comes to solving equations, if we can determine where our function is equal to 0, that is, we can find its x intercepts and we can find its discontinuities, we're actually in a very comfortable place where we can solve equations, especially as one presses forward into calculus. I should also mention that when f of x is itself a ratio, the x intercepts are going to be the values that make the numerator go to 0, right? So you have this fraction, whatever makes the top go to 0, right? That's going to give you 0 still. And then when the denominator goes to 0, that's going to give you a discontinuity. It's either a removable discontinuity because the top also went to 0 or it's a vertical asymptote. So we have fractions. We care about what makes the numerator go to 0. What makes the denominator go to 0? Because if we can find the x intercepts, then we can solve any equation, but it's also going to be important to find discontinuities as well. What's outside the domain of the function, okay? So consider the following. f of x equals 4 thirds x to the 1 third times 2x plus 1 plus 2x to the 4 thirds. And so if we were to try to solve this equation equal to 0, we would factor it, right? How do we factor that? Well, the most important thing to look for when you're factoring is the idea of a common divisor, greatest common divisor. What are the common divisors between these things, okay? And so when looking at this thing right here, what I see when I'm looking for a GCD, I'll notice that if you look at the coefficients, I have a 4, I have a 2. You can't take more away than what the least holding term has. So if you look at 4 and 2, they both have a factor of 2, because 4 itself could factor as 2 and 2. So I could rewrite this thing as 2 times 2x to the 1 third, 2x plus 1, plus 2x to the 4 thirds here, like so this is over 3. For which then we can identify the 2 that can factor out, so we take it out. And this gives us 2 times 2 thirds, x to the 1 third, times 2x plus 1, plus x to the 4 thirds. So you can take out the coefficient of 2, that's one thing to do. Another thing we want to do when we factor is we don't want fractions if we can avoid it. So you'll notice that the first term has a 1 third as a factor, right? The other one doesn't, but you can always insert a 1 third by just times it by a strategic number 1. If you just times it by, you know, 3 over 3, then you could factor out the 1 third, which is common to both. This will make it a life a little bit easier for us by taking out the fraction. So I'm actually going to take out a 2 thirds, and that'll leave behind. You'll see x, or 2 times x to the 1 third, times 2x plus 1, and then you're going to get 3 times x to the 4 thirds. So as you're factoring out your GCD, I would recommend clearing denominators in the process. That's going to be helpful for us. So that we took care of the coefficients. That's helpful. What we really want to grab is going to be the powers of x. Now, when you look at the powers of x, one thing I should mention is that here you notice you have this product, x to the 1 third and 2x plus 1. When you're trying to factor, multiplying is the exact opposite of what you're trying to do. It's kind of inhibiting your ability to factor. So don't distribute this thing. That's a no-no. Don't do it. Leave it factored. Alright, so you'll notice here that you have this x to the 1 third, this x to the 4 thirds right here. When you're trying to factor out powers of x, you have to select the smallest power. And so when you're taking the minimum of 1 third and 4 thirds, the smallest power is going to be 1 third. So we're going to factor 1 third away from both of these terms. So it turned out our GCD was 2 thirds x to the 1 third. That's what we want to factor out. We want to factor out a 1 third here. But how do you factor out 1 third away from the 4 thirds right here? Well, remember factoring division or basically the same thing. When you have something like x to the m plus n power, remember this is the same thing as x to the m times x to the n. So we can factor using exponents here. So why that's helpful is that x to the 4 thirds is the same thing as x to the 1 third times x to the 3 thirds. Like so 1 and 3 gives us 4. And this x to the 1 third we factor away. So that then gives us this factored form 2 thirds x to the 1 third. That was our GCD which we factored out. What was left behind? We get 2 times 2x plus 1. And then you're going to get 3 times x to the 3 thirds. But the 3 thirds is just the first power. So I'm just going to write that as 3 times x. Now you'll see that the GCD is taken out. There's nothing else we can factor amongst all these things. Now at this moment, feel free to distribute and combine like terms. This then gives us 2 thirds x to the 1 third. We're going to get 4x plus 2 plus 3x. And so simplifying that we end up with 2 thirds x to the 1 third times a 4. Sorry, we're going to get a 7x plus 2 when we add those all together. So now once our function is completely factored. We're now in a situation we find this x there says when this thing equals 0. Applying the 0 product property, we have that either. We have that either x to the 1 third equals 0. Which if you take the cube of both sides, you see x equals 0. Or we have that 7x plus 2 equals 0. For which you subtract 2, you get 7x equals negative 2. You divide by 7, you get x equals negative 2 sevenths. And so we see then the x intercepts are going to be x equals 0 and x equals negative 2 sevenths. We do have to pay attention to the domain, right? Or are there any values of x that were forbidden? For which when we look at the original function, always look to the original function here. The only thing that gives us any concern is we do have fractional powers, which means radicals, but we're taking cube roots. There's no even roots, no square roots whatsoever. So the domain of this thing, the domain of F is going to be all real numbers. So since there's no restriction, there's no discontinuities. 0 and negative 2 sevenths are our two x intercepts for this function. Let's take a look at another example of this type of principle here. So if we were to look at this, what can we factor away? I like to look at the coefficients, right? You have 12, which factors as 3 and 4. We have a negative 9, which factors as 3 and 3. So our GCD is going to grab that factor of 3 right there. But what else can we grab, right? Notice we have an x minus 2 to the 4 thirds power and we have an x minus 2 to the 1 third power. So again, taking the minimum of 4 thirds and 1 third, this is going to give you 1 third. So we're going to slap on an x minus 2 to the 1 third power. You can only take away the smallest power of x minus 2 right here. So we're going to factor that out as well. And so in factoring, we're going to take out the 3 times x minus 2 to the 1 third. What's left behind? Well, when we factor the 3 away from the 12, we get a 4. So factoring and division are really just the same thing, right? 12 divided by 3 is 4. Then if we take x minus 2 to the 4 thirds power, if we factor away the third power, we're going to get x minus 2 to the 1 third power, right? You subtract the exponents 4 thirds minus 1 third is equal to 3 thirds, which is equal to 1 like we saw in the previous example. So this actually just will become x minus 2. And then with the next part, you have 9. We factor out 3. 9 divided by 3 is 3. And then we have x minus 2 to the 1 third power divided by x minus 2 to the 1 third power. When you factor, you're dividing, these things will actually just cancel each other out. So you actually just get a 3 when you're done. Like so. For which we took out the GCD. There's nothing else we can factor. So we're going to multiply these things out. So just distribute the 4. We get 3 times x minus 2 to the 1 third power. We're then going to get 4x minus 8 minus 3, for which I can combine that constant and we're going to get that G of x here is equal to 3 times x minus 2 to the 1 third power. And then we get 4x minus 11, for which again, there's no discontinuities here. The domain of G is going to be all real numbers. Because again, the only thing that potentially could have been a problem was this fractional exponent, but an odd denominator means cube root here. There's no problem with domain. So as we use the zero product property, setting these things equal to zero, right? We see that either x minus 2 to the 1 third power equals 0, for which take the cube of both sides, we get x minus 2 equals 0. Adding 2, we get x equals 2 as 1 of the roots. So let's record that over here, x equals 2. This one comes about when we take 4x minus 11 equals 0. Add 11 to both sides. You get 4x equals 11. Divide both sides by 4. You get x equals 11 fourths. And that then gives us the other x intercept, which is in the solution to this equation. G of x equals 0. All right. Let's do another example. This one we have f of x equals 1 third times 2x to the negative 1 third minus 5 times x to the 2 thirds. We already have a 1 third factored out. Don't distribute it. We want things to be factored. When you look at the coefficients 2 and negative 5, there's no common devices to pull up there. So let's just look at the powers of x. If you take the minimum power this time, the first x contributes a negative 1 third. The second x contributes a 2 thirds. The minimum is actually negative 1 third because negatives are smaller than positives. So you're going to factor out a negative power here, okay? So in doing that, h of x is going to become 1 third times x to the negative 1 third. So then you're going to get 2 times x to the negative 1 third over x to the negative 1 third. And then you're going to have minus 5 times x to the 2 thirds minus x to the negative 1 third. The first one's pretty easy because you have x to the negative 1 third divided by itself. So let's just go down to a 1. So you end up with 1 third x to the negative 1 third times just a 2. But for the next one, we're going to subtract powers, right? So you're going to get negative 5 x to the 2 thirds minus a negative 1 third, right? Since you're subtracting a negative, you're actually adding a power. And so we end up with here 2 minus 5 times x to the 3 thirds power, like so. I should also mention that if you have a negative x one, negative x one actually means take reciprocals, right? x to the negative 1 third is the same thing as 1 over x to the 1 third. And since we already have a fraction of 1 third, we can actually put those things together. We get 3 times x to the 1 third, like so. Now 3 thirds, of course, is just a 1. So simplifying our function h right here, h of x is going to look like we have a 2 minus 5x on the top. And then we have a 3x to the 1 third in the denominator. So in this regard, if we set the numerator equal to 0, right? Because when this thing equals 0, only the numerator has to be 0. So you take 2 minus 5x equals 0. That will give us 2 equals 5x. Divide both sides by 5. You get x equals 2 fifths. So this is our x-intercept. Our x-intercept is going to equal 2 fifths. Now unlike the previous examples, you'll see in this one that there actually is a value that makes the denominator go to 0. That is, there is some value that's undefined. In this situation, if we set the denominator equal to 0, divide both sides by 3. You get x to the 1 third equals 0. Take the cube of both sides. That is the third power. You get x equals 3. And so we see that this, there actually is a problem with the domain. This thing has a vertical asymptote at x equals 0. So there is a discontinuity. Now 0 and 2 fifths don't overlap, of course, here. So we see that there's an x-intercept at 2 fifths, but we have a vertical asymptote at 0. That is the domain of the function as everything except for 0 here. Let's look at just a couple more examples here. So let's time, let's throw some exponentials into the play. We've been only doing power functions so far. f of x here is 2x times e to the x plus x squared times e to the x. So what are our common divisors here? What's our GCD in this situation? When you look at the coefficient, you have 2 versus 1. Nothing we can take out. If you look at x and x squared, we take away the smallest power. That's the first power. We're going to take away an x. And then you look at it in terms of the exponentials. They're both divisible by e to the x for which you're going to get an e to the x right there. So that's what we need to factor out from our function f of x here. So f of x, if we take out the GCD, we're going to end up with x times e to the x. That was the GCD. Then the first one, if you take away x and e to the x, you're left with just a 2, right? For the second one, if you take away x, you have just an x because you took 1x away from the x squared. And you also take away the e to the x. So you just end up with an x right there. In terms of the domain of this thing, it's going to be all real numbers. Power function. There's no restriction there. e to the x is all real numbers. If we set this equal to 0, by the 0 product property, we have three cases to consider. We have x equals 0. We have e to the x equals 0. And we have 2 plus x equals 0, which implies that x equals negative 2. For which we grab this one. No big deal. We grab this one. No big deal. On the middle case, though, I should mention that e to the x can't equal 0. The range of e to the x is only pods of numbers. e to the x cannot equal 0. And if you don't remember that, to solve the equation, you take the natural log of both sides. You'd get that x equals the natural log of 0, which is actually like negative infinity. That's a vertical acetone. It's the same problem. You don't get a solution in that situation. The 0 product property says that if a product of things is equal to 0, then one of the factors must have been 0. It doesn't tell you that every factor has to be 0 for some choice of x. e to the x actually doesn't give you a solution in that situation. So in terms of the x intercepts of this function, they occur at x equals 0 and negative 2. Like so. Let's do one more example. This time, let's involve a logarithm into play here. So what can we do here? Well, the first thing that catches my attention is we have division by x, right? That does affect the domain of things, right? What's the domain of this thing? So x can't equal 0, but we also have a logarithm. x can't be 0 or a negative. So the domain of g is going to be 0 to infinity. We see that. So with the domain now established, I notice I have an x cubed divided by x. So that would simplify to be 3x squared natural log of x minus x squared. Like so. And then as I start searching for the gcd, I notice that both of the terms are divisible by x squared, right? You have an x squared right here and right here. So let's factor it out. And so we see that g of x is equal to, we'll take out the x squared, right? x squared times 3 natural log of x minus 1, like so. x squared equals 0. So the first factor, x squared equals 0, you take the square root of both sides, you're going to get x equals 0. Okay? The other one, we're going to take 3 times the natural log of x minus 1 is equal to 0, like so. And then if we start solving this, we're going to add 1 to both sides. Add 1, add 1. We end up with 3 times the natural log of x is equal to 1. Next, we're going to divide both sides by 3. Like 3. In which case, then we get the natural log of x equals 1 third. In which case now, we have to move the natural log to the other side. It switches to the natural exponential. And we end up with x equals e to the 1 third power, or if you prefer the cube root of e as the number there. And so then our x intercepts for this function, the x intercepts are going to be at x equals 0 and the cube root of e. Now, we do have to throw out 0, right? Oh, yeah, because 0 is outside the domain of this thing. So maybe we didn't notice it until the end, that's okay. So we actually need to remove it from consideration. And we get that this function actually only has one x intercept. And this happens at the cube root of e. Now there's some more examples I want to do of solving these equations, you know, setting it equal to 0 and even factoring. But this, what we've seen so far here is illustrate just how useful GCDs are. Factoring out the Graves common divisor is a very, very important tool, especially as one steps into calculus. Because I have a little secret for you. All five of these examples, a, b, c, d, e that we've done in this video are calculus problems, right? I mentioned earlier this idea, you have to solve for this so-called derivative is equal to 0. You might not know what a derivative is yet, that's okay. But what I want to mention to you is that each and every one of these functions that we're dealing with right here is a derivative of a function. This is a calculus problem, you do that calculation. Then the next step of the problem is once you calculate the derivative, you have to set it equal to 0 and solve, right? You have to figure out what's outside the domain and what's equal to 0. So these are actually calculus problems without the calculus step, hence why we called them precalculus. So I can't emphasize enough how important this GCD factoring is because there will be a litany of calculus problems you will see in the future, which factoring by GCD alone will be sufficient to help you solve these problems and find the so-called critical numbers. That is the numbers that make the derivative equal to 0 and those numbers outside the domain of the function.