WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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We are going to continue our lesson on polynomials by discussing roots and zeroes.
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The number of roots can be determined by looking at the degree of the polynomial equation,
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because a polynomial equation of degree n has n roots.
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This tells you only the number of roots; it doesn't tell you about the nature of the roots; the roots may be real or complex.
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For example, if I am given a polynomial such as 5x⁶ + 2x³ - x² + 9x - 2 = 0,
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a polynomial equation, I see that the degree equals 6; this tells me that there are 6 roots for this equation.
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Now, they can be real; and for example, these roots could be something like 1, -2...those are rational numbers.
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They actually could be irrational numbers, like the square root of 3.
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You also can have complex numbers as roots.
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Recall that complex numbers have two parts--something like 2 + 3i; and the first part here is real, and the second part is imaginary.
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And these are in the form a + bi.
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We are going to talk, in a few minutes, about how these complex roots occur as conjugate pairs.
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For example, 2 + 3i and 2 - 3i would be a conjugate pair--the same values here, but opposite signs; or 4 + i and 4 - i.
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So, I know that here I have 6 roots; but I have no idea if they are real, rational, irrational, or complex, or what combination of those.
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However, I can determine at least some of that through using Descartes' Rule of Signs.
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And there are two sections to this rule: the first section helps you determine the possible numbers of positive real roots.
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Then, we will talk about determining the possible number of negative real roots.
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So, first we are just looking at the positive real roots: if we let p(x) be a polynomial with real coefficients (no imaginary coefficients--
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just real coefficients), arranging in descending powers, recall that descending powers would be something like
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f(x) = x⁶ + 5x⁵ + 7x⁴ - x³ + 6x² - 2x + 4.
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This is a degree 6 polynomial; and I have 6, 5, 4, and on down.
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If you are trying to work with Descartes' Rule of Signs, the first thing to do is check the polynomial.
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And if it is not arranged in descending powers, you need to put it that way.
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OK, so I have this arranged in descending powers; the number of positive real roots
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is the number of changes in sign of the coefficients, or is less than this by an even number.
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Let's look at what that means: I want to look for the number of sign changes of the coefficient.
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So, here I have...this coefficient is a 1, and that is positive; 5 is positive; 7 is positive; -1--so there is a sign change.
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This is positive; this is negative; so that is one sign change.
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OK, now here, I am going from -1 to positive 6; again, I have a sign change.
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Here, I am going from a positive to a negative--another sign change; negative to positive--another sign change.
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That gives me 1, 2, 3, 4--4 sign changes means there are 4 positive real roots, or less than this by an even number.
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So, less than 4 by an even number would mean 4 - 2 (would give me 2), or another even number, 4: 4 - 4 would give me 0.
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There are 4, 2, or 0 positive real roots.
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So again, take the polynomial; arrange it in descending powers; and then look at the number of sign changes.
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I have 1, 2, 3, 4 sign changes; that tells me that I will have, at most, four positive real roots.
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However, I may have less than this by an even number (4 - 2: 2; 4 - 4; 0).
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So, I have three possibilities: I may have 4 positive real roots, 2 positive real roots, or 0 real roots.
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And I have, since the degree equals 6, a total of 6 roots.
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I have a total of 6; of these, 4, 2, or 0 may be positive real roots.
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OK, the second part of this is looking at the number of negative real roots.
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The number of negative real roots is the number of changes in sign of the coefficients of the terms p(-x), or is less than this by an even number.
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Let's continue on with the example that we just looked at, where we were given
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f(x) = x⁶ + 5x⁵ + 7x⁴ - x³ + 6x² - 2x + 4.
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Now, I found there were four sign changes, which means 4, 2, or 0 positive real roots.
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For the negative real roots, I have to look at f(-x).
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So, I need to change these x's to -x and be very careful with the signs; OK.
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So, this is going to give me a coefficient here of -1.
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But if I take a negative to an even power, it is going to become positive.
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-1 to the sixth power is going to become positive, so this is going to give me f(-x); here it is just going to be x⁶.
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Here, I have -1 times x; you can look at it that way--it is -1 times x⁵.
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Well, if I take a -1 to an odd power, it is going to remain negative; so, -x⁵ times 5 is going to give me -5x⁵.
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Here, I have a negative coefficient to an even power; it is going to become positive, so this is really 7x⁴.
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Here, I have a negative coefficient to an odd power, so it will remain negative; so this is -x³,
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but it is times a negative: so -x³ times -1 becomes + x³.
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OK, I have a -x; this actually should be outside...(-x)² is going to give me -x times -x; that is going to give me a positive.
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So, this is going to be plus 6x².
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Here, I have -x times -2; that is +2x; and my constant remains positive.
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OK, number of sign changes: I am going to look for the changes in sign.
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This is positive out here; so a positive to a negative--that is 1; a negative to a positive--that is 2.
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This stays positive, positive, positive, positive; the number of sign changes equals 2.
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OK, so the number of negative real roots is the number of changes in sign of the coefficients of the term p(-x), or less than this by an even number.
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Therefore, I am going to have two negative real roots, or less than this by an even number.
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Well, 2 minus 2 is 0; I can't go any lower than that for the number of roots; so there are 2 or 0 negative real roots.
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Again, this power is 6; I have 6 total roots; I have 2 or 0 negative real roots; and last slide, we talked about having 4, 2, or 0 positive real roots.
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So, we covered the real roots; there also may be complex roots, so let's talk now about the total roots and the different combinations that you could have.
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We can use Descartes' Rule of Signs to determine the possible combinations of the real and complex roots.
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So, in that example above, f(x) = x⁶ + 5x⁵ + 7x⁴ - x³ + 6x² - 2x + 4,
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the total roots (this is real and complex) equal 6.
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Positive real roots: using Descartes' Rule of Signs, I found that I could have 4, 2, or 0.
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Negative real roots: I found f(-x) and looked for the sign changes, and found two of those.
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So, the number of negative real roots could be 2 or 0.
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Now, I can figure out the combinations: I know that they need to total 6, and I have
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my positive real roots' possibilities and my negative real roots; and the last part of this is complex roots.
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So, I have positive real roots, the number of negative, and the number of complex; and these need to total 6; the total must equal 6.
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So, I said that, for positive real roots, I could have 4; then, I may have 2 negative real roots.
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4 and 2 is 6, so that leaves me with no complex.
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I could have 4 positive real roots; I could have 0 negative real roots; 4 and 0 is 4; to total 6, I will have to have 2 complex roots.
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OK, another possibility: I have 2 positive real roots and 2 negative real roots: 2 and 2 is 4; 2 more complex roots will give me 6.
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Or I could have 2 positive real roots and 0 negative real roots.
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2 and 0 is 2; to total 6, I will have 4 complex roots.
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OK, finally, I may have 0 positive real roots and 2 negative real roots.
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0 and 2 is 2, so I have to have 4 complex roots.
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Then, I may have 0 positive real roots and 0 negative real roots, and that gives me 0 and 0.
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So, to total 6, I would have to have 6 right here.
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And since 3 times 2 is 6, I expect 6 combinations, and that is what I have here.
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So, Descartes' Rule of Signs, and knowing that the degree is 6 (so I have 6 total roots)
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allows me to figure out the possible combinations of the roots of a polynomial.
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Conjugate roots: we talked a bit about complex conjugates and the fact that there are complex roots.
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But getting into a bit more detail: if the polynomial p(x) has real coefficients...
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the coefficients are not imaginary, which is what we will be working with--real coefficients--for now...
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the complex roots of p(x) occur as complex conjugates.
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And you saw this earlier on, when we worked with quadratic equations and quadratic functions.
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Thinking about something like this: x² + 4 = 0, if I wanted to find the solutions for this, I could say, "OK, x² = -4."
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Now, using the square root property, I take the square root of both sides; and this gives me x = ±√-4.
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Recall that the square root of -1 equals i; then I can rewrite this as x = ±√-1, times √4,
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or x = ±i√4, or x equals ±2i.
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And we can look at this, also, in a different way: we can say x = 0 + 2i, or x = 0 - 2i.
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And you can then see that this is a pair of complex conjugates.
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And it is because of this, where we are finding the square root, that we end up with plus or minus some number.
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And therefore, these complex roots of polynomials occur as complex conjugates.
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Another example of complex conjugates would be, say, 8 + 5i and 8 - 5i--the same values, but switch the sign--or 2 + 7i and 2 - 7i.
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And this explains why, when we talked about the number of negative real roots and positive real roots,
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we said it was that number (say, 4), or less than that by 2.
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And the reason it goes down by pairs is because the complex conjugates occur in pairs.
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So, they could take up two of the spots for the roots; and so they are going to decrease the number of the real roots by 2 or multiples of 2.
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Here, we are asked to solve this polynomial equation, x³ - x² - 6x = 0.
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I am going to solve this by factoring; and the first step is to factor out the greatest common factor.
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And here, I see I have a common factor of x; there is an x in each of these.
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So, factor that out; this is going to give me x times x², minus x, minus 6, equals 0.
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Don't forget to bring this x along as you factor this, because this is going to give us one of the solutions.
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I have x² - x - 6 that I need to factor; this is in the form (x + a constant) (x - a constant), because the sign is negative.
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Factors of 6: 1 and 6, 2 and 3; and I need these to add up to -1, and their signs are going to be opposite.
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These (2 and 3) are close together, so if I make the larger one negative and the smaller one positive, I am going to get a -1.
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Therefore, 3 is negative, and 2 is positive.
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Now, according to the zero product property, if any of these terms (x, x + 2, or x - 3) is 0, then this product will be 0.
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And it will equal the right side of the equation.
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Therefore, x equals 0 (and make sure you don't leave this one out, or you will be missing one of the solutions), x + 2 = 0, or x - 3 = 0.
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So here, I don't have to do anything further with this.
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I just have that one of the solutions here is x = 0.
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Here, I have to subtract 2 from both sides to get x = -2; and here, I need to add 3 to both sides.
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So, solutions: x = 0, x = -2, and x = 3; these are all solutions to this equation.
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And I solved this by factoring out the greatest common factor, x, factoring this trinomial,
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and then using the zero product property to solve for x in each of these expressions' terms.
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Here, I am asked to solve x⁴ - 256 = 0.
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This is actually the difference of two squares; this is in the form a² - b².
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So, it is going to factor out to (a + b) (a - b).
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And if you look at it and think about it this way, x², squared, is x⁴.
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And if you look at this one, 256, and take the square root of that, it is actually 16.
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Therefore, this is telling me that a equals x², and b equals 16.
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So, I can factor it as follows: a + b (that is x² + 16), times a - b (or x² - 16).
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OK, and looking at what I have, I can't do anything else with x² + 16.
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But I recognized again, here: I have the difference of two squares; this time, a equals x, and b equals 4.
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So, it is going to factor out to (x + 4) (x - 4); and these are set equal to 0.
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Now, I use the zero product property, which is going to tell me that I could have x² + 16 = 0, x + 4 = 0, or x - 4 = 0.
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And let's work with these simpler ones first.
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Simply subtract 4 from both sides to give me x = -4.
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Add 4 to both sides: x = 4; I have two of my solutions.
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Now, looking over here, it is a little bit more complex: subtract 16 from both sides--that gives me x² = -16.
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Now, I am going to take the square root of both sides, and you can see that this ends up being ±√-16.
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And this is a negative number; and since I know that √-1 equals i, I can rewrite this as x = ±i√16.
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Well, the square root of 16 is 4, so this gives me a complex conjugate pair, plus or minus 4i.
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So, I have four solutions: x = 4i, x = -4i, x = 4, and x = -4.
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And let's just think about Descartes' Rule of Signs and show that it predicted the possibilities for the type of roots that I could get.
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Since I have x⁴ - 256, if I look at the number of sign changes for this, this is positive to negative (one sign change).
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This tells me that I am going to have one positive real root, or less than that by an even number;
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but I can't go there, because then I would be going into negative numbers,
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and I can't say there are -1 real roots; that wouldn't make sense.
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So, it is just one positive real root.
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Now, looking at f(-x): this gives me -x⁴ - 256.
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Well, this -1, when you take it to the fourth power, is just going to become positive; so this gives me this.
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And again, I have one sign change; so this tells me that I have one negative real root.
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Since the degree here is 4, I have 4 total roots.
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So, this is going to leave me with one positive real, one negative real, and two complex roots,
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which is exactly what I see: a positive real, a negative real, and the set of complex conjugates.
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Determine the possible combination of positive real roots, negative real roots, and complex roots.
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We will use Descartes' Rule of Signs to determine this, and we will start out by thinking about the total.
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Since the degree is 3, there are 3 total roots.
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Now, using the rule of signs, I am going to look for f(x) and the sign changes: this is 2x³ - 3x² + 4x - 5.
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Number of sign changes: well, this is positive, and this is negative--that is one.
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I am going from negative here to positive here; that is two; from positive to negative--that is 3.
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So, the number of sign changes equals 3.
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Therefore, the number of positive real roots is 3, or less than this by an even number (3 - 2 is 1).
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You can't go any lower than that; if I subtracted by 2 again, I would get a negative number.
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So, I have either 3 or 1 positive real roots.
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All right, now let's look at the negative scenario for f(-x) to figure out the negative real roots.
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2 times -x cubed, minus 3 times -x squared, plus 4 times -x, minus 5:
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OK, this is going to give me f(-x) =...this is going to remain negative, so this is going to give me -2x³.
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A negative and a negative (squared) is going to give me a positive, so this will become x².
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This is negative here, though, so it is -3x².
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4 times -x is -4x, minus 5.
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OK, the number of sign changes: none, none, none: the signs are all negative--the coefficients of f(-x).
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So, the number of sign changes equals 0; so there are 0 negative real roots.
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OK, so let's figure out what we have going on here.
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We have positive real roots, negative real roots, and complex.
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And remember: these need to total 3.
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Positive real: I could have 3; negative real: 0; I need them to total 3--this already does, so the complex is going to be 0, since this totals 3.
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OK, another possibility is that I have one positive real root, 0 negative real roots.
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1 and 0 is 1; I need it to total 3; so there must be 2 complex real roots.
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So, the possible combination of positive real roots, negative real roots, and complex roots is 3, 0, 0, or 1, 0, 2.
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And I know I have three total roots, because the degree is 3.
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I use Descartes' Rule of Signs to tell me that I had either 3 or 1 positive real roots; I have 0 negative real roots.
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And then, figuring out what is lacking to get my total of 3, I could fill it in with complex roots.
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OK, determine the possible combinations of positive real roots, negative real roots, and complex roots.
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Degree equals 4, so I have four total roots.
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f(x) = -3x⁴ - 4x³ + 2x² + 6x + 7.
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So, positive real roots: let's look for sign changes between these coefficients.
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A negative to a negative; a negative to a positive (that is 1); a positive to a positive--no sign change; a positive to a positive again.
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So, the positive real roots equals 1; and I can't go by less than that,
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because if I subtracted 2, I would go into negative numbers.
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So, the number of positive real roots is 1.
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Now, looking at negative real roots: I am going to take f(-x) to get
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-3 times -x to the fourth, minus 4 times -x to the third, times 2 times -x squared, plus 6 times -x, plus 7.
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OK, this gives me f(-x) =...well, -1 to the fourth power--this will become positive; so I have -3x⁴
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Here, -x cubed...this is going to remain negative as a coefficient...times -4; that is going to become positive, so this is going to give me + 4x³.
00:29:07.700 --> 00:29:20.000
-x squared is going to give me positive x² times 2; so that is + 2x²; 6 minus -x is -6x; plus 7.
00:29:20.000 --> 00:29:32.600
OK, negative real roots is going to be determined by the sign changes of the coefficients of f(-x), according to Descartes' Rule of Signs.
00:29:32.600 --> 00:29:44.600
A negative to a positive; that is one sign change; that stays positive; a positive to a negative: 2; a negative to a positive: 3.
00:29:44.600 --> 00:29:56.100
So, here I have 3 or less than that by an even number: 3 - 2 is 1.
00:29:56.100 --> 00:29:59.200
You can't go any less than that.
00:29:59.200 --> 00:30:10.500
I have a total of 4 roots; I have one positive real root, and either 3 or 1 negative real roots.
00:30:10.500 --> 00:30:25.600
So, let's look at the possibilities: positive real roots, negative real roots, and complex.
00:30:25.600 --> 00:30:37.800
Positive real: 1; then I could have a negative real root totaling 3: 1 + 3 is 4; I only have 4; so complex must be 0.
00:30:37.800 --> 00:30:46.800
Or I could have 1 positive real root and 1 negative real root; and I am going to total those to get 2.
00:30:46.800 --> 00:30:55.000
But I should have 4; so in this case, there would be two complex roots--a pair of complex conjugates.
00:30:55.000 --> 00:31:00.900
OK, so I determined there were four total roots, because the degree here is 4.
00:31:00.900 --> 00:31:09.100
Using Descartes' Rule of Signs, the positive case, f(x), I found that there is one positive real root.
00:31:09.100 --> 00:31:14.800
f(-x) gave me three sign changes, so there are either three or one negative real roots.
00:31:14.800 --> 00:31:23.800
And then, to get a total of four, I had zero complex roots in this case, and two in this case.
00:31:23.800 --> 00:31:27.000
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