WEBVTT mathematics/math-analysis/selhorst-jones
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Welcome back to Educator.com.
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Today, we are going to talk about roots (also called zeroes) of polynomials.
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We briefly went over what a root is when we talked about the properties of functions.
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But let's remind ourselves: the zeroes of a function, the roots of an equation, and the x-intercepts of a graph are all the same thing.
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They are inputs (which are just x-values) where the output is 0; it is where it comes out to be 0--things that will make our expression give out 0.
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The roots, zeroes, x-intercepts of f(x) = x² - 1 and y = x² - 1 are x = -1 and x = +1,
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because those two values, f(x) and y, are equal to 0.
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If we plug in -1, (-1)² will become positive 1, minus 1 is 0.
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If we plug in positive 1, 1² is 1, minus 1 is 0.
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Those two things cause it to give out 0, and the roots of the polynomial are the x-values where the polynomial equals 0.
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The roots of the polynomial x² - 1 are -1 and positive 1.
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Being able to find roots in functions is important for many reasons; and it will come up very often when you are working with polynomials.
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If you continue on to calculus, you will see how roots can be useful for finding lots of information about a function.
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So, it is very important to have a grasp of what is going on there and be able to find roots.
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Roots in graphs: if you have access to the graph of a function/equation, it is very easy to see where the roots are.
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Of course, you might not see precisely, because it is a graph, after all, and it might be off by a little bit.
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But you can get a very good sense of where they are: it is where the graph cuts the horizontal axis--the x-intercepts.
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Why? Because here we have f(x) = 0 or y = 0, depending on if it is a function or an equation.
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Since that means our height is at 0 there, then every place where we cross the x-axis must be a root--it's as simple as that.
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This also gives us a nice mnemonic to remember what the word "root" means--
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it can be a little hard to remember the word "root," since we aren't that used to using it.
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But we could remember it as where the equation or the function grows out of the x-axis--where it is 0.
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It is like it is the ground; think of it as a plant rooted in the ground.
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A function or equation has its roots in the x-axis; a tree has its roots in the earth, and a function has its roots in a height of 0.
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So, a root is where it is growing up and down; it is where it is held in our plane, held in our axes.
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And that is one way to remember what a root is.
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How do we find the roots of a polynomial? Well, at first we might try a naive approach and attempt to solve the way we are used to.
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Naive is just what you have done before--what seems to make sense--without ever really having had a whole lot of experience about it.
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So, the naive attempt would be probably to just isolate the variable on one side.
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That is what we did with a bunch of other equations before, so let's do it again.
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Now, in some cases, this will actually work, and we will find all of the real solutions.
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For example, if we have f(x) = x - 3, then we set it equal to 0, because we are looking for the roots; we are looking for when 0 = x - 3.
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We move that over, and we get x = 3; great.
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Or if we had y = x³ + 1, we set that to 0, and we have 0 = x³ + 1,
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because we are looking for what x's cause us to have 0 = x³ + 1.
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So, we have x³ = -1; we take the cube root of both sides, and we get x = ³√-1, which is also just -1.
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So, in both of these cases, the naive method of isolating for variables worked just fine.
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But that is definitely not going to be the case for all situations.
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The naive method of isolation will fail us quite quickly, even when used on simple quadratic polynomials.
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Consider f(x) = x² - x - 2--that is not a very difficult one.
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But this method of trying to isolate will just fail us utterly if we use it here.
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So, 0 = x² - x - 2...we might say, "Well, let's get the numbers off on one side."
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We have 2 = x² - x; but then, we don't have just x; so let's pull out an x; we get 2 = x(x - 1).
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And well, we are not really sure what to do now; so let's try another way.
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0 = x² - x - 2; let's move x over, because we are used to trying to get just x alone.
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So, we have x here; but then we also have x² here; so let's divide by x.
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We get 1 = (x - 2)/x; once again, we are not really sure what to do.
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Let's try again: 0 = x² - x - 2, so let's move over everything but the x².
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Maybe x² is the problem; so we will get x + 2 = x².
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We take the square root of both sides; we remember to put in our ± signs: ±√(x + 2) = x.
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I don't really know how to figure out what x's go in there to make that true.
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So, in all three of these cases, it is really hard to figure out what is going on next.
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If we are going to try to isolate, we are going to get these really weird things.
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This method of isolation that we are used to isn't going to work here, because we can't get x alone; we can't get the variable alone on one side.
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It is not going to let us find the roots of polynomials if we try to isolate.
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But at least we could trust it in those previous examples; we saw that we can trust it when it does work.
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So, we might as well try it first--no, it is even worse: the naive method of isolation
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can make us miss answers entirely, even though we think we will know them all.
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So, we will think we have found the answers; but in reality, we will only have found part of the answers.
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Consider these two ones: if we have 0 = x² - 1, we move the 1 over; and then we take the square root of both sides.
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The square root of 1 is 1; the square root of x² is x; great.
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For this one over here, we have 0 = p² + 3p, so we realize that we can divide both sides by p.
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And since 0/p is just 0, we have 0 = p²/p (becomes p); 3p/p becomes 3; so we have 0 = p + 3.
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We move the 3 over; we get p = -3; great--we found the answers.
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Not quite: those above things are solutions, but in each case, we have missed something.
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We have been tricked into missing answers by trying to follow this naive method.
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The other solutions for this would be x = -1 and p = 0, respectively.
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The mistakes that we forgot were a ± symbol over on this one; we forgot to put a ± symbol when we took the square root;
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and then, the other one was dividing by 0, because when we divided, we inherently forgot
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about the possibility that, if we were actually dividing by 0, we couldn't divide by 0.
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So, those are the two mistakes; but even if you personally wouldn't have made those same mistakes,
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this example shows how it is easy to forget those things in the heat of actually trying to do the math.
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You might forget about that; you might accidentally make one of these mistakes; so it is risky to try this method of isolation.
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We need something that works better.
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We find the roots of the polynomial by factoring; we break it into its multiplicative factors.
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Let's look at how this works on the example that we couldn't solve with naive isolation.
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We have f(x) = x² - x - 2; we have 0 = x² - x - 2, because we are looking for when f(x) is equal to 0.
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And then, we say, "Let's factor it; let's break it into two things."
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So, we have (x - 2) and (x + 1); and if we check, that does become it: x times x becomes x²; x times 1 becomes + x;
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-2 times x becomes - 2x; -2 times 1 becomes -2; so yes, that checks out to be the same thing as x² - x - 2.
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0 = x - 2 and 0 = x + 1 is how we now set these two things equal to 0; we have 0 = x - 2 and 0 = x + 1.
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And we get x = 2 and x = -1, and we have found all of the solutions for this polynomial.
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Why does this work, though? We haven't really thought about why it works.
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And we don't want to just take things down and automatically say, "Well, my teacher told me that, so that must be the right thing."
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You want to understand why it is the right thing.
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Teachers can be wrong sometimes; so you want to be able to verify this stuff and say, "Yes, that makes sense,"
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or at least have them explaining and saying, "Well, we don't understand quite enough yet;
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but later on you will be able to see the proof for this"; you really want to be able to believe these things,
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beyond just having someone tell you by word of mouth.
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So, to figure out why this has to be the case, we will consider 0 = ab.
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The equation is only true if a or b, or both of them, is equal to 0; if neither a nor b is equal to 0, then the equation cannot be true.
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If a = 2 and b = 5, then we get 10, which is not equal to 0.
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As long as a or b is 0, it will be true, because it will cancel out the other one.
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But if both of them are not 0, then it fails, and it is not going to be 0.
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It is the exact same thing happening with x² - x - 2; we have 0 = x² - x - 2,
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which we then factor into (x - 2) and (x + 1); so let's use two different colors, so we can see where this matches up.
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We are pairing this to the same idea of the a times b equals 0; it is (x - 2) (x + 1) = 0.
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The only way that this equation can be true is if x - 2 = 0 or x + 1 = 0.
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Just as we showed up here, it has to be the case that a or b equals 0 for that to be true.
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So, it must be the case that either (x - 2) or (x + 1) equals 0, if this is going to be true.
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So, our solutions are when either of the two possibilities is true--if the possibility is true, if one of them is true, then the whole thing comes out.
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So, either case being true makes it acceptable; that gets us 2 (keeping with our color coding) and -1 as the two possibilities.
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So, by breaking x² - x - 2 into its factors, we can find its roots.
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So, this is how we find polynomial roots, in general: the first thing we do is set the whole thing as 0 = polynomial.
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We have to have some polynomial, and then it is 0 equals that polynomial.
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The next thing we do is factor it into the smallest possible factors; we break it down into multiplicative factors.
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And then finally, we set each factor equal to 0, and we solve for each of them.
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So, in step number 2, we are going to get things like 0 = (x + a)(x + b)(x + c)(x +d) and so on, and so on, and so on.
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And then, in step 3, we set each factor to 0; so we get things like x + a = 0, at which point we can solve and say, "Oh, x = -a."
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That is one of our possible solutions; and from there, you can work out all of the roots of the polynomial.
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Caution--this is very important: notice that it is extremely, extremely important to begin by setting the equation as 0 = polynomial.
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I have seen lots of mistakes where people forgot to set it as 0 = polynomial.
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If it isn't, if it was something like 5 = (x - 2)(x + 1), we can't solve for the solutions from those factors.
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Those factors are now meaningless; they aren't going to help us.
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We need the special property that 0 turns everything it multiplies into 0.
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Without that special property, this method just won't work.
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Consider if we had something like 5 = ab; there is no way that we could just figure out what the answers are here.
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It is not just simply that a has to be 0 or b has to be 0, because a could be 5 and b could be 1.
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Or b could be 5 and a could be 1, or a could be 25 and b could be 1/5; or a could be 100 and b could be 1/20.
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We have lots of different possibilities--a whole spectrum of things; there are way too many possible solutions.
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We need that special property of 0 = ab to be able to really say, "That thing is 0, or that thing is 0"; that is what we know for sure.
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That is how we get useful information out of it.
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That is why it has to be 0 = polynomial; if you don't set it up as that before you try to factor it,
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before you try to do the other steps, you are just not going to be able to get the answer,
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because we need that special property that 0 has when it multiplies other things.
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0 multiplying something automatically turns it to 0; if we don't have that special property, things just won't work.
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Factoring is not necessarily easy: say we have something like
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x⁵ + 6.5x⁴ - 17x³ - 41x² + 24x, and we want to know what the zeroes are.
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Well, if we knew its factors, we would be able to break it into (x + 8)(x + 2) times x times (x - 0.5) times (x - 3).
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And it would be really easy to figure out what the polynomial's roots are.
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At this point, we say, "Well, great; x + 8 becomes x = -8; x + 2 becomes x = -2; x + 0 becomes x = 0;
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x - 0.5 becomes x = positive .5; and x - 3 becomes x = positive 3--great; I found it; it is really easy to find its roots."
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But how do you factor a monster like that?
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Ah, there is the problem: factoring can be quite difficult.
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Luckily, by this point, you have been certainly practicing how to factor for years in your algebra classes.
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By now, you have done lots of factoring; you are used to this; you have played with polynomials a bunch in previous math classes.
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And all that work has a use, and it is here, finding roots; we can break things down into their factors and find roots.
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But there is no simple procedure for factoring polynomials.
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Once again, remember: if you were confronted by something like this, you would probably have a really difficult time
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figuring out what its factors were--figuring out how you can break that down into factors.
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There is no simple procedure for factoring polynomials.
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High-degree polynomials can just be very difficult to factor; happily, we are not going to really see such polynomials.
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Most courses on this sort of thing don't end up giving you very difficult things to factor at this stage.
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So, we won't really have to worry about factoring really difficult polynomials.
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We will be able to stick to the smaller things.
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So, let's have a quick review of how you factor small things like quadratics.
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We are going to see a bunch of quadratics; they are very important--they pop up all the time in science.
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So, let's look at a brief review of how to factor a quadratic polynomial.
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Remember, quadratic is a degree 2, and a trinomial just means 3 terms.
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So, if we have a quadratic trinomial in its normal form, then we have ax² + bx + c.
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If we want to factor that, we will turn it into a pair of linear binomials: degree 1 and 2 terms (linear and binomial).
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We would want to break this into (_x + _) (_x + _).
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Great; to be equivalent to the above, the coefficients of x must give a product of a.
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So, a has to come out with the red dot here, times the red dot here.
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And the constants have to give a product of c; so the blue dot here times the blue dot here has to come out to be c.
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Also, b has to add up from the products of the outer blanks and the inner blanks.
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So, it has to be that the red dot here times the blue dot here, plus the blue dot here times the red dot here, comes up to be this b here.
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So, it is mixed out of the two of them.
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Don't worry if that is a little bit confusing right now; you will be able to see it as we work through examples.
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The b has to add up to the products of the outer blanks and the inner blanks.
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The a has to come from the first blanks, and the c has to come from the last blanks.
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Don't worry about memorizing this, though--it is just a sense of what is going on in practice.
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Let's look at an example: we want to factor 2x² - 5x - 12.
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So, we know, right away, that we want to break it into the form (_x + _)(_x + _).
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The first thing we notice is that we have this 2 at the front; and 2 only factors into 2 times 1.
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We can't break it up into anything else really easily; so let's put 2 times 1 down as 2x times x.
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We have to put the 2 somewhere; so it is either going to be (2x + _)(1x + _), or it is going to be the 2 over here and the 1 here.
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It doesn't really matter what order we put it in; so we will put the 2 at the front.
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We have (2x + _)(x + _); now, what is going to go into those other blanks?
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Now, we need to factor the -12; so let's factor 12 first: we notice that 12 can break into 1 times 12, 2 times 6, or 3 times 4.
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And one of the factors has to be a negative, because we have a negative in front of the 12.
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So, they have to be able to multiply to make a -12; so there is going to have to be a negative on either the 1 or the 12.
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One of those two will have to have a negative on it; we don't know which one, but it is going to be one of them.
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Or it is going to be -2, or -6; and then finally, for the last pair, it would be -3 or -4.
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I am not going to use all of these at once; we will have to figure out which one is right.
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But one of them will have to be negative, because of this negative sign up here.
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We start working through this; and we know that there is this 2x here at the front.
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We have this 2x at the front; so it is going to multiply this one, and it is going to effectively double whatever we put here.
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So, the difference between one of the numbers doubled and its sibling (the one here times this one in front of it) must be -5, because we get -5x.
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We notice that 3 - 2(4)...2 times 4 is 8; 3 is 3; so the difference between those is 5.
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So, we can set it up as 3 - 2(4) = -5; and we get (2x + 3)(x - 4); and we have factored this out; we have been able to work it out.
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Now, there are various ways, various tricks that have been taught to you in previous classes.
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But the important thing is just to set up and have an expectation of what form you are trying to get.
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And then, plug things in and say, "Yes, that would work; that would get me what I am looking for" or
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"No, if I plug that in, that won't work; that won't get me what I am looking for."
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As long as you work through that sort of thing, you will be able to find the answer eventually.
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It is always a good idea, though, to check your work; you will find the answer, but it is really easy to make mistakes.
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So, even on the easiest of problems (like the one we were just working on), there are lots of chances to make mistakes; trust me.
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I make mistakes; everybody makes mistakes; the important thing is to catch your mistakes before they end up causing problems.
00:17:18.600 --> 00:17:24.200
So, it means that you should always try expanding the polynomial after you have factored it, to make sure you factored correctly.
00:17:24.200 --> 00:17:28.000
And it is OK to do this in your head; once you get comfortable with doing this sort of thing
00:17:28.000 --> 00:17:33.300
(and by now, honestly, you probably have had enough experience with this that you can just do this in your head reasonably quickly),
00:17:33.300 --> 00:17:37.800
it is OK to do it in your head; the important part is that you want to have some step where you are checking back on what you are doing.
00:17:37.800 --> 00:17:42.900
So, either do it on the paper (if it is a long one) or do it in your head (if it is something that is short enough, easy enough, for you to check).
00:17:42.900 --> 00:17:45.100
But you want to make sure that you are checking your work.
00:17:45.100 --> 00:17:50.900
For example, if we have 2x² - 5x - 12, we figured out that that breaks into (2x + 3) and (x - 4).
00:17:50.900 --> 00:17:53.200
But we want to check and make sure that it is right.
00:17:53.200 --> 00:18:03.300
So, we check and make sure: 2x + 3 times x - 4...we get 2x², and then 2x(-4), 3x...we get -8x + 3x...
00:18:03.300 --> 00:18:11.900
3 times -4...we get -12; we combine our like terms of -8x + 3x, and we get 2x² - 5x - 12.
00:18:11.900 --> 00:18:20.400
Sure enough, it checks out; we have what we started with, so we know that our factoring was correct; we did a good job.
00:18:20.400 --> 00:18:27.300
Factoring higher-degree polynomials: in general, factoring polynomials of any degree is going to be similar to what we just did on these previous few slides.
00:18:27.300 --> 00:18:32.000
The only difference is that it will become more complex as they become longer, as we get to higher and higher degrees.
00:18:32.000 --> 00:18:37.800
So, for example, if we had something like a cubic--if we had ax³ + bx² + cx + d--
00:18:37.800 --> 00:18:42.700
we would probably want to set it up as (_x + _)(_x² + _x + _).
00:18:42.700 --> 00:18:50.900
If we are going to be able to break it up and factor it, it is going to have to factor into these two things, (_x + _)(_x² + _x + _).
00:18:50.900 --> 00:19:01.100
Notice also that this is a degree 1, and this is a degree 2; and when you multiply these two together, you will get back to a degree 3 over here.
00:19:01.100 --> 00:19:05.500
Adding up the degrees on the right side has to be what we had on the left side.
00:19:05.500 --> 00:19:08.900
We could also work on quartic, a degree 4; and we might try one of these two templates.
00:19:08.900 --> 00:19:12.900
If we have ax⁴ + bx³ + cx² + dx + e, we might break it
00:19:12.900 --> 00:19:23.100
into (_x + _)(_x³ + _x² + _x + _); or we might break it into (_x² + _x + _)(_x² + _x + _).
00:19:23.100 --> 00:19:28.100
And so on and so forth...clearly, this is going to become more difficult as we get to higher and higher degrees.
00:19:28.100 --> 00:19:33.500
The higher the degree of a polynomial, the more complicated our template is going to have to be for where we are going to fit things in.
00:19:33.500 --> 00:19:37.400
The more choices we are going to have, the more difficult it is going to be to do this.
00:19:37.400 --> 00:19:42.400
Luckily, we are only occasionally going to need to factor cubics, things like this where it is degree 3.
00:19:42.400 --> 00:19:45.800
And we are very, very seldom going to see anything of higher degree.
00:19:45.800 --> 00:19:48.500
So, don't worry too much about having really difficult ones.
00:19:48.500 --> 00:19:56.000
But just be aware that factoring really large, high-degree polynomials can actually be pretty difficult to do.
00:19:56.000 --> 00:20:01.500
Roots imply factors: we have this useful trick if we want to break out these higher polynomials.
00:20:01.500 --> 00:20:05.800
If we know one of a polynomial's roots, we automatically know one of its factors.
00:20:05.800 --> 00:20:09.600
Remember: one use of finding the factors of a polynomial is to find its roots.
00:20:09.600 --> 00:20:16.500
If you find a factor of the form (x - a), then you set that equal to 0, and you know that that is going to be x = a.
00:20:16.500 --> 00:20:25.100
It turns out that the exact opposite is true; if we know a polynomial has a root at x = a, then it also means we have a factor of x - a.
00:20:25.100 --> 00:20:31.900
So, if we have a root x = a, then that turns into a factor, x - a, just because of this equation here,
00:20:31.900 --> 00:20:36.300
where we are setting that factor equal to 0, which gives us the root.
00:20:36.300 --> 00:20:42.500
We won't prove this; it requires a little bit of difficult mathematics, and some things that we actually haven't covered in this course yet.
00:20:42.500 --> 00:20:50.500
But we can see it as a theorem: Let p(x) be a polynomial of degree n; then if there is some number a,
00:20:50.500 --> 00:20:57.500
such that p(a) = 0 (that is to say that a is a root of our polynomial, p--if we plug a in, we get 0),
00:20:57.500 --> 00:21:04.400
then there is some way to break it up so that it is p(x) = (x - a), our factor (x - a) that we know
00:21:04.400 --> 00:21:11.900
from our root at x = a, times q(x), where q is some other polynomial of degree n - 1,
00:21:11.900 --> 00:21:25.100
because this here is degree 1; so when we multiply it by a degree n - 1, we will be back up to our degree n polynomial that we originally had.
00:21:25.100 --> 00:21:31.600
If we manage to find one or more roots, sometimes the problem will give them to use; other times, we will get lucky, and we might just guess one.
00:21:31.600 --> 00:21:35.000
This theory means we automatically know that many factors of the polynomial.
00:21:35.000 --> 00:21:40.300
For example, say we know that p(x) = x³ - 2x² - 13x - 10.
00:21:40.300 --> 00:21:45.300
And we are told that p(5) = 0; we know that 5 is a root.
00:21:45.300 --> 00:21:52.000
Then, we automatically know that at x = 5 we have a zero; so (x - 5) is our root.
00:21:52.000 --> 00:21:57.400
We plug that in; we know it is going to be (x - 5)(_x² + _x + _).
00:21:57.400 --> 00:22:04.900
Now, we don't know what is in these blanks yet; we still don't know what is going to be in there.
00:22:04.900 --> 00:22:13.700
But we are one step closer to figuring out what those factors are, for being able to figure out what has to go in those blanks.
00:22:13.700 --> 00:22:16.600
Later on, in another lesson, we will use this fact to great advantage--
00:22:16.600 --> 00:22:19.800
this fact that knowing the root automatically means you know a factor--
00:22:19.800 --> 00:22:24.000
When we learn about the intermediate value theorem to help us find roots,
00:22:24.000 --> 00:22:31.800
and the polynomial division using those roots to break down large polynomials into smaller, more manageable factors.
00:22:31.800 --> 00:22:38.400
Not all polynomials can be factored, though; even with all of this talk of factoring polynomials, there are some that cannot be factors,
00:22:38.400 --> 00:22:43.400
not because it is difficult or really hard to do, but because it is just simply impossible.
00:22:43.400 --> 00:22:49.700
Consider this polynomial: f(x) = x² + 1; if we try to find its roots, then we have 0 = x² + 1.
00:22:49.700 --> 00:22:56.700
So, we have x² = -1; but there is no number that exists that can be squared to become a negative number.
00:22:56.700 --> 00:22:59.800
No number can be squared to become a negative number; why?
00:22:59.800 --> 00:23:06.600
Consider: if we have (-2)², that becomes positive 4; if we square any negative number, it becomes a positive number.
00:23:06.600 --> 00:23:10.900
If we square any positive number, it stays positive; if we square 0, it stays 0.
00:23:10.900 --> 00:23:15.300
So, there is no number that we have, that we can square, and get a negative number out of it.
00:23:15.300 --> 00:23:21.300
Thus, the polynomial has no roots; and since it has no roots from that theorem we just saw, it can't have any factors.
00:23:21.300 --> 00:23:27.500
So, it has no roots; therefore, it cannot be reduced into smaller factors.
00:23:27.500 --> 00:23:33.600
And something that cannot be reduced, we call irreducible; it is not reducible.
00:23:33.600 --> 00:23:37.900
Now, I will be honest: what I just told you isn't really the whole story.
00:23:37.900 --> 00:23:41.800
More accurately, we can't factor all polynomials *yet*.
00:23:41.800 --> 00:23:51.800
The previous slide that we just saw is perfectly true, but only if you are working with just the real numbers (which have the symbol ℝ like that).
00:23:51.800 --> 00:23:54.900
Now, that is what we are normally working with; so it is kind of reasonable to say this.
00:23:54.900 --> 00:23:59.100
But it turns out that there is a hidden type of number that we haven't previously explored.
00:23:59.100 --> 00:24:01.300
You might have seen this in previous math classes, even.
00:24:01.300 --> 00:24:09.300
We will learn about the complex numbers later on; complex numbers can give us a way to factor these supposedly irreducible polynomials.
00:24:09.300 --> 00:24:14.500
So, they are irreducible for real numbers; but they are not irreducible for complex numbers.
00:24:14.500 --> 00:24:19.700
Now, we will learn about them in the lesson that is named after these numbers--our lesson on complex numbers.
00:24:19.700 --> 00:24:24.500
But for now, we are just working with real numbers; and in general, we will just be working with real numbers in this course.
00:24:24.500 --> 00:24:27.300
Real numbers are really useful; you can do a lot of stuff with them.
00:24:27.300 --> 00:24:30.100
So, it is enough for us to be working with real numbers, generally.
00:24:30.100 --> 00:24:34.500
That means, for us right now, at least some polynomials are simply irreducible.
00:24:34.500 --> 00:24:39.900
And we can't always find roots for everything in a polynomial, because we can't break it down,
00:24:39.900 --> 00:24:44.300
because there are things that just don't have roots, based on how real numbers work.
00:24:44.300 --> 00:24:47.700
Now, we will talk about complex numbers later on; but I just wanted to point this out--
00:24:47.700 --> 00:24:52.200
that I am not telling you the whole story right now, because we don't want to get confused with complex numbers.
00:24:52.200 --> 00:24:58.300
But for our purposes right now, with real numbers, there are some things that are simply irreducible.
00:24:58.300 --> 00:25:01.600
There is a limit to how many roots or factors a polynomial can have.
00:25:01.600 --> 00:25:04.800
Now, roots and factors are basically two sides of the same thing.
00:25:04.800 --> 00:25:15.700
Since x = a as a root is the same thing as knowing (x - a) as a factor, they are just two sides of the same thing.
00:25:15.700 --> 00:25:18.200
So, we will consider them as roots/factors of polynomials.
00:25:18.200 --> 00:25:26.700
A polynomial of degree n can have, at most, n roots/factors; so we can have a maximum of n of these.
00:25:26.700 --> 00:25:34.900
Why is this the case? Well, consider this: every factor comes in the form (x + _), or even larger if the factor is irreducible.
00:25:34.900 --> 00:25:40.100
It might be (_x² + _x + _); but the very smallest has to be (x + _).
00:25:40.100 --> 00:25:46.900
If we break a polynomial into its factors, we are going to get (x + _) (x + _)...so on and so on, up until (x + _).
00:25:46.900 --> 00:25:55.500
Now, if we had more than n factors, then that would mean that we have (x + _) multiplying by itself more than n times.
00:25:55.500 --> 00:26:03.700
So, if we have x multiplying more than n times, then it has to have a degree larger than n; its exponent is going to have to be greater than that.
00:26:03.700 --> 00:26:15.800
If we wanted to max out at x², but we had (x + _)(x + _)(x + _)...well, that is going to become x³
00:26:15.800 --> 00:26:21.000
plus stuff after it, which is not going to be x²...not going to be degree 2.
00:26:21.000 --> 00:26:29.200
So, if we have a degree n polynomial, the most factors we can possibly have are n factors, n roots,
00:26:29.200 --> 00:26:35.200
because otherwise, we would have too many factors, and we would blow out the degree of the polynomial.
00:26:35.200 --> 00:26:41.000
Thus, the most roots/factors a polynomial can have is equal to its degree.
00:26:41.000 --> 00:26:45.200
We also can get information about the possible shape of a polynomial's graph from its degree.
00:26:45.200 --> 00:26:50.000
A polynomial of degree n can have, at most, n - 1 peaks and valleys.
00:26:50.000 --> 00:26:52.700
Formally speaking, that is relative maximums and minimums.
00:26:52.700 --> 00:26:58.100
For example, if we have x⁴, then that means we have n = 4 (our degree).
00:26:58.100 --> 00:27:10.400
So, n - 1 = 3; so we look over here, and we have one valley here, one peak here, one valley here.
00:27:10.400 --> 00:27:22.600
This is also a relative minimum, a relative minimum (that is what we mean by "valley), and a relative maximum (that is what we mean by "peak").
00:27:22.600 --> 00:27:30.300
So, n - 1 tells us the most bottoms and tops we can have, before they either go off to positive infinity or go off to negative infinity.
00:27:30.300 --> 00:27:36.000
Now, we can't prove this here, because it requires calculus; but it is connected with the maximum number of roots in a polynomial.
00:27:36.000 --> 00:27:41.000
And if you go on to take calculus (which I heartily recommend), you will very clearly, very quickly see it.
00:27:41.000 --> 00:27:45.300
It becomes very clear in calculus; it is one of the important points of what you do in calculus.
00:27:45.300 --> 00:27:50.000
So, you will think, "Oh, that makes a lot of sense," because the possible peaks and valleys
00:27:50.000 --> 00:27:55.500
are connected to a polynomial that has a degree that is 1 less; and that is why it is connected.
00:27:55.500 --> 00:28:01.100
Don't worry about it too much right now; but it is very interesting, and very obvious, if you go on to take calculus.
00:28:01.100 --> 00:28:06.800
Notice that, in both of the previous properties, it was described as "at most."
00:28:06.800 --> 00:28:13.200
Just because a polynomial has degree n does not mean it will have n distinct roots or n - 1 peaks and valleys.
00:28:13.200 --> 00:28:17.500
We aren't necessarily going to have to have that many; it is just that we can have up to that many.
00:28:17.500 --> 00:28:24.300
Consider f(x) = x⁵ + 1; this graphs like this, but from this graph, we can see clearly:
00:28:24.300 --> 00:28:28.000
we only have one root, and we have no peaks or valleys.
00:28:28.000 --> 00:28:33.400
The degree gives an upper limit on how many there can be, but it doesn't tell us how many there will be.
00:28:33.400 --> 00:28:39.200
It just that the maximum is this; but you could definitely have fewer.
00:28:39.200 --> 00:28:41.100
All right, we are ready for some examples.
00:28:41.100 --> 00:28:46.000
The first one: we want to find the zeroes of f(x) = 3x² - 23x + 14.
00:28:46.000 --> 00:28:51.100
So, this is a textbook example--literally a textbook example, since this is effectively a textbook.
00:28:51.100 --> 00:28:55.500
So, you plug in 0, because we are looking for when f(x) is equal to 0.
00:28:55.500 --> 00:29:11.400
0 = 3x² - 23x + 14; we know we are going to be looking for 0 =...something where it is going to be (_x + _)(_x + _).
00:29:11.400 --> 00:29:18.000
What are we going to slot in there? Well, we notice that here is 3; the only way we can break up 3 is 3 times 1.
00:29:18.000 --> 00:29:22.300
There are no other choices; so we either have to have 3 go for the first x or 3 go to the second x.
00:29:22.300 --> 00:29:31.800
So, let's set it as 0 = (3x + _); and we will have 1x, so just x, plus _.
00:29:31.800 --> 00:29:37.100
Great; now, at this point, we also say, "We have 14 over here; how can 14 break up?"
00:29:37.100 --> 00:29:43.000
Well, we can have 1 times 14, or we can have 2 times 7; those are the only choices.
00:29:43.000 --> 00:29:45.800
So, we are going to have to plug in either 1 times 14 or 2 times 7.
00:29:45.800 --> 00:29:49.200
But now, we also have to take this -23 into consideration.
00:29:49.200 --> 00:29:52.800
If we have -23, then we are going to have at least one negative over here.
00:29:52.800 --> 00:29:56.100
And since it comes up as a positive, it is going to have to be that they are both negatives.
00:29:56.100 --> 00:30:02.500
So, one of them is negative, so they are going to both be negatives; so it will be -1(-14) or -2(-7).
00:30:02.500 --> 00:30:07.100
So, -1 times -14...we will notice that, either way we put that in, that won't work out.
00:30:07.100 --> 00:30:22.900
But we can plug in -2 times -7, and we can amp up this -7; so + -7 here...put in the -2 here...and we get 0 = (3x - 2)(x -7).
00:30:22.900 --> 00:30:27.700
We have managed to factor it; let's really quickly check what we have here--does this work out?
00:30:27.700 --> 00:30:47.400
Check (3x - 2)(x - 7); we would get 3x² - 21x - 2x + 14, so 3x² - 23x + 14; it checks out; sure enough, it is good.
00:30:47.400 --> 00:30:55.000
So, at this point, we break this down into two different possibilities: either 3x - 2 = 0, or x - 7 = 0.
00:30:55.000 --> 00:31:01.000
So, 3x - 2 = 0 or x - 7 = 0 are the two different worlds where this will be true,
00:31:01.000 --> 00:31:04.600
where we will have found a root where the whole expression will be equal to 0.
00:31:04.600 --> 00:31:20.700
So, 3x - 2 = 0: we get 3x = 2; x = 2/3; over here, x - 7 = 0; we have x = 7; so our answers are x = 2/3 and x = 7; those are the roots for this polynomial.
00:31:20.700 --> 00:31:27.700
Great; if f(2) = 0, factor f(x) = x³ - 7x + 6.
00:31:27.700 --> 00:31:43.900
Remember: if we know that at x = 2 we have a zero (at x = 2 there is a root), then that means there is a factor in that polynomial of (x - 2).
00:31:43.900 --> 00:31:52.100
How do we figure that out? Well, we notice that x = 2; then it is x - 2 = 0, so that implies that it has to be a factor of (x - 2) in there.
00:31:52.100 --> 00:31:57.100
We can use that piece of information; we know that f(x) is going to have to break down with an (x - 2) in there.
00:31:57.100 --> 00:32:04.400
So, let's set it up like normal: 0 = x³ - 7x + 6; but what we just figured out here...we know that there is a factor of (x - 2).
00:32:04.400 --> 00:32:15.500
So, we can also write this as 0 = (x - 2)(_x² + _x + _); what are going to go into those blanks?
00:32:15.500 --> 00:32:20.700
Well, at this point, we just use a little bit of logic and ingenuity, and we can figure this out.
00:32:20.700 --> 00:32:24.200
Well, we know that...what is in front of this x³? It is effectively a 1.
00:32:24.200 --> 00:32:32.300
So, if there is a 1 in front, we have x times x²; whatever goes into this blank is going to determine what coefficient is in front of it.
00:32:32.300 --> 00:32:37.300
So, since we want a 1, it has to be that there is a 1 here, as well.
00:32:37.300 --> 00:32:48.000
What about the very end? Well, the only thing that is going to create the ending constant is going to be the other constants.
00:32:48.000 --> 00:32:54.900
So, the constants that we have here are -2 and whatever goes into that blank; so it must be that -2 times _ here becomes 6.
00:32:54.900 --> 00:33:00.100
-2 times -3 becomes 6; so we have a -3 here.
00:33:00.100 --> 00:33:04.000
And finally, what is going to go into this blank here?
00:33:04.000 --> 00:33:11.700
We think about this one, and we know that we want to have 0x² come out of this; there are no x²'s up here.
00:33:11.700 --> 00:33:21.500
So, we have + 0x²; so whatever we put into this blank must somehow get us a 0x² to show up.
00:33:21.500 --> 00:33:24.700
So, x times x² is x³; so we are not going to worry about that.
00:33:24.700 --> 00:33:30.100
But x times _x is going to be x²; let's do a little sidebar for this.
00:33:30.100 --> 00:33:43.600
x times _x will become _x²; and -2 times...we already filled in that blank...1x² is going to be -2x².
00:33:43.600 --> 00:33:51.600
Now, we want the 0x² out of it; so it must be that, when we add these two things together, it comes out to be 0x².
00:33:51.600 --> 00:33:55.000
What does this have to be? It has to be positive 2x².
00:33:55.000 --> 00:34:02.300
We know that positive 2x², minus 2x², comes out to be 0x²; so it must be that this is a positive 2x.
00:34:02.300 --> 00:34:13.500
So, we write this whole thing out: 0 = (x - 2)(x² + 2x - 3); and we have been able to figure out that that works.
00:34:13.500 --> 00:34:27.100
We check this out and do a really quick check; so x³ + 2x² - 3x - 2x² - 4x - 6...
00:34:27.100 --> 00:34:32.200
x³ checks out; 2x² - 2x² cancel each other; that checks out.
00:34:32.200 --> 00:34:38.400
-3x - 4x; that becomes -7x, so that checks out; -6...that checks out as well.
00:34:38.400 --> 00:34:47.600
Great; we have a correct thing, so 0 = (x - 2)(x² + 2x - 3) is correct.
00:34:47.600 --> 00:34:54.800
We factored it properly; so at this point, the only thing that we have left to factor is the x² + 2x -3.
00:34:54.800 --> 00:35:03.500
0 = (x - 2)(_x + _)(_x + _); what goes in those first blanks?
00:35:03.500 --> 00:35:08.400
Well, we just have a 1 in front of that; so it is 1 and 1...we don't have to worry about it that much.
00:35:08.400 --> 00:35:17.500
What else is going to go in there? Well, -3 is at the very end; we have +2x, so it must be that the negative amount is smaller than the positive amount.
00:35:17.500 --> 00:35:27.900
So, it is going to be + -1 and + 3; -1 times 3 gets us -3, and everything else checks out.
00:35:27.900 --> 00:35:36.400
x times x is x²; plus 3x minus x...that gets + 2x; and minus 1 times 3 gets us -3; so that checks out.
00:35:36.400 --> 00:35:38.500
We did another check in our heads really quickly.
00:35:38.500 --> 00:35:51.800
So finally, we have 0 = (x - 2)(x - 1)(x + 3); we break this up into three different worlds;
00:35:51.800 --> 00:36:06.700
set each world equal to 0: x - 2 = 0; x - 1 = 0; x + 3 = 0; so we have x = 2; x = 1; x = -3.
00:36:06.700 --> 00:36:11.700
Those are all of the roots for this polynomial.
00:36:11.700 --> 00:36:17.000
All right, the next example: give a polynomial with roots at the indicated locations and the given degree.
00:36:17.000 --> 00:36:25.800
Now remember, a root can become a factor; so if we know that we have a root at -3, then that becomes a factor of...
00:36:25.800 --> 00:36:36.500
if it was x = -3, then it becomes (x + 3); if we had x = 8, then that would become (x - 47)...
00:36:36.500 --> 00:36:41.600
I'm sorry; I accidentally read the wrong one--read forward one; that is (x - 8).
00:36:41.600 --> 00:36:47.300
And then finally, if we had x = 47, we would have (x - 47) as our factor.
00:36:47.300 --> 00:36:57.000
So, those three things together...we have (x + 3)(x - 8)(x - 47), and that right there is a polynomial.
00:36:57.000 --> 00:37:05.700
We know it has degree 3, because we have x times x times x; that is going to be the largest possible exponent we can get on our variable.
00:37:05.700 --> 00:37:10.200
That will come out to be x³, so we have a degree 3; and we know it has roots in all of the appropriate places.
00:37:10.200 --> 00:37:14.500
And we are done--that is it; we could expand this, and we could simplify, if we wanted.
00:37:14.500 --> 00:37:19.100
We weren't absolutely required to by the problem, and this is a correct answer.
00:37:19.100 --> 00:37:27.700
It is a polynomial; it is not in that general, standard form that we are used of _x to the exponent, _x to the exponent, _x to the exponent.
00:37:27.700 --> 00:37:32.300
But it is still a polynomial, so it is a pretty good answer; we will leave it like that.
00:37:32.300 --> 00:37:41.000
The next one: we have -2 and positive 2, so we have (x + 2) for the -2 and (x - 2) for that.
00:37:41.000 --> 00:37:48.600
Why? x = -2; we move that over; we get x + 2 = 0; x = +2...we move that over and we get x - 2 = 0.
00:37:48.600 --> 00:37:55.100
So, we get (x + 2)(x - 2); but if we multiply those two together, we just have a degree of 2, and we want a degree of 4.
00:37:55.100 --> 00:38:04.400
So, we need to somehow get the degree up on this thing, but have the same roots--not have to accidentally introduce any more roots.
00:38:04.400 --> 00:38:09.000
So, if we introduced multiplying just by x twice, we would have introduced a root at x = 0.
00:38:09.000 --> 00:38:16.400
So, we can't just do that; but we do realize that if we just increase this to square it on both of them, they will still have the same roots.
00:38:16.400 --> 00:38:22.400
It is just duplicate roots showing up; so (x + 2)² + (x - 2)²...we have hit that degree of 4,
00:38:22.400 --> 00:38:25.100
because each one of these will now have a degree of 2.
00:38:25.100 --> 00:38:35.600
Alternately, we could have done this as (x + 2) to the 1, (x - 2) cubed, or (x + 2)³(x - 2)¹.
00:38:35.600 --> 00:38:40.100
Any one of these would have degree 4, and have our roots at the appropriate place.
00:38:40.100 --> 00:38:47.400
Great; the final example: What is the maximum possible number of roots and peaks/valleys for each of the following polynomials?
00:38:47.400 --> 00:38:59.400
So, for our first one, f(x), we notice that this has a degree of 3; so n = 3 means the maximum number of roots it can have is 3,
00:38:59.400 --> 00:39:10.400
and the maximum peaks/valleys is one less; n - 1 is 3 - 1, is 2.
00:39:10.400 --> 00:39:14.100
So, the maximum number of roots is 3; the maximum peaks/valleys is 2.
00:39:14.100 --> 00:39:19.300
We don't necessarily know it will have that many; all we know is that that is the maximum it could possibly have.
00:39:19.300 --> 00:39:29.600
The next one: we notice that the degree for this one is 47; so if n = 47, then the maximum roots are going to be equal to that degree.
00:39:29.600 --> 00:39:39.200
The maximum peaks/valleys are going to be one less than that degree, so we will get 46, one less than that.
00:39:39.200 --> 00:39:48.200
The final one: for this one, we think, "Oh, 10³, so it is 3!"--no, we have to remember that this is not a variable.
00:39:48.200 --> 00:39:53.900
This here is a variable; so it is x¹, so its degree is just 1.
00:39:53.900 --> 00:40:05.400
For that one, degree 1...we will change over to the color green...n = 1 means the maximum roots are just 1;
00:40:05.400 --> 00:40:15.400
and the maximum peaks/valleys are going to be one less than 1, so 1 - 1 = 0.
00:40:15.400 --> 00:40:20.400
Now, why is that the case? Well, think about it: 10³ - 5...10 cubed is just some constant;
00:40:20.400 --> 00:40:29.300
it happens to be 1,000, but that is not really the point; so 1000x - 5 is just going to be a very steep line.
00:40:29.300 --> 00:40:35.000
x - 5...it will intersect here; but does it ever go up and down--does it ever undulate in weird ways?
00:40:35.000 --> 00:40:41.600
No, it never does anything; we just have a nice, straight line, since it is a linear thing (linear like a line).
00:40:41.600 --> 00:40:45.600
So, since it is a linear expression, it never undulates--never has any peaks or valleys.
00:40:45.600 --> 00:40:50.800
So, it never has any relative maximums, and no relative minimums; and that is why we have 0 there--it makes sense.
00:40:50.800 --> 00:40:54.300
All right, I hope everything there made sense; I hope you got a really good understanding of roots,
00:40:54.300 --> 00:40:57.500
because roots will come up in all sorts of places; they are really important to understand.
00:40:57.500 --> 00:41:01.300
It is really important to understand this general idea, because you will see it in other things, being changed around.
00:41:01.300 --> 00:41:05.800
But if you understand this general idea, you will be able to understand what is going on in later things and different courses.
00:41:05.800 --> 00:41:07.000
All right, see you at Educator.com later--goodbye!