WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we will be discussing the rational zero theorem.
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What this theorem tells you is the possible rational zeroes of a polynomial.
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And again, it just helps you generate a list of possibilities.
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So, if f(x) is some polynomial function with integer coefficients, then you can generate a list of possible rational zeroes
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by looking at two things: the leading coefficient and the constant.
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If p/q is a fraction in simplest form that is a zero of f(x), then p is a factor of a₀,
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which is the constant; and q is a factor of a < font size="-6" > n < /font > , which is the leading coefficient.
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So, let's think about what this is saying.
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p/q: here, p is a factor of the constant; so the numerators of all the elements of the lists that we are going to generate consist of factors of the constant.
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The denominators consist of factors of the leading coefficient.
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Using this equation, using the rational zero theorem, what I can do is make a list of the factors of the constant,
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make a list of the factors of the leading coefficient, and then find all possible combinations,
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using those factors as the numerator (the factors of the constant) and the factors of the leading coefficient as the denominator.
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Find all the possible combinations; and then, that will give me a list of possible rational zeroes.
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Now, this is only a list; and we are going to talk about how you check the list and determine which ones,
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if any, are actually zeroes; because it may turn out that none of these are zeroes.
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It may turn out that the zeroes are irrational or complex numbers.
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But if there are rational zeroes, they will be members of this list.
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OK, using an example: f(x) = 4x² + x - 3.
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Here, my leading coefficient is 4, and the constant is 3.
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Now, I am going to first just think about the factors: the factors of 4 are going to be 1, 2, and 4.
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And these could be positive or negative; and I am going to worry about the signs in a minute.
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Right now, I am just going to look for the factors.
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The factors of 3 are 1 and 3; now, these factors of the leading coefficient are going to comprise the denominator.
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So, these are possibilities for the denominator.
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Here, these factors of the constant are going to be possibilities for the numerator.
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So, I am going to take each one; I am going to go in an organized manner;
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I am looking for different possibilities for p/q, for the factors of 3 over the factors of 4.
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p/q could be...I could have 1 in the numerator and 1 in the denominator, and it could be positive or negative,
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because the factors here could include -1 times -4 (that would give me 4), and so on.
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So, I could have all different combinations of positive and negative.
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OK, I could have 1 for the numerator and 2 for the denominator, plus or minus.
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I could have 1/4; I could have 3/1, plus or minus; I could have 3/2, plus or minus; and plus or minus 3/4.
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And then, I am just writing this in simpler form: this is just ±1, ±1/2, ±1/4; this is just 3 (3/1 is just 3); ± 3/2; ±3/4.
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So, this gives me the list of possible rational zeroes.
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Now, let's factor this out and see what happens, just to check and make sure that what we come up with actually is on this list.
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Let's take the corresponding equation and find the zeroes.
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4x² + x - 3 = 0: you can actually factor this out, and this is going to come out to (4x - 3) (x + 1).
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So, this would be 4x², and this would be 4x - 3x (would give me x); -3 times 1 is -3.
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Using the zero product property, I am going to get 4x - 3 = 0 and x + 1 = 0.
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So, this is going to give me 4x = 3, or x = 3/4; this will give me x = -1.
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I look, and I have 3/4 on the list, and I have -1 on the list; both are on the list.
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So, what this gave me is a list of 1, 2, 3, 4, 5, 6 possibilities, positive or negative: that is 12 possibilities.
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And I found two of them; I was able to find this by factoring.
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But what it is showing is that these were on the list; but again, this is just a list of possibilities.
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And you need to be able to figure out which possibilities are correct.
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So, what we are going to do is generate lists of possibilities, and then talk about
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what you can do with that information--how you can find which possibilities are correct.
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OK, if the leading coefficient of the polynomial is equal to 1, then the rational zero theorem tells us that any rational zero of f(x) is a factor of the constant.
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Let's think about what this means: we said that the list of rational zeros equal p/q,
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where p is factors of the constant--let's just say the constant--and q are factors of the leading coefficient.
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Well, if the leading coefficient is 1, then this becomes p/1, so it is just going to be factors of the constant.
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So, in the situation where the leading coefficient is 1, you just have to look at the factors of the constant.
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And those will give you your list of possible rational zeroes.
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For example, if I have f(x) = x⁴ - 3x³ + 2x² + x - 8.
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I have a leading coefficient of 1, so I don't need to worry about that: p/q = p/1 = p.
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Now, here p is going to equal the factors of the constant, which will be factors of 8, equal ±1, ±2, ±4, ±8.
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Now, I mentioned that these are just possibilities; so I need to check these.
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And if you will recall, we can check to see if something is a factor by using synthetic division.
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We also talked about how, if a number is a 0, then if you take the value of the function,
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you are going to find that if f(a) = 0, then x - a is a factor.
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So, for example, if x - 2 is a factor of f(x), then f(2) will equal 0.
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And recall that using synthetic division, we can find the value of a function by looking at the remainder.
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So, what we can do is go back to synthetic division (or synthetic substitution, as we call it in this case)
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and divide the polynomial by x - a, where a is one of the factors we are checking.
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And if the remainder is 0, then we know we have a factor or a rational zero, in this case, if the remainder is 0.
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So, these factors are also zeroes, because f(a) is 0.
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So, if I go ahead and divide, for example, f(x) by x - 2, that will allow me to check to see if 2 is a factor.
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If 2 is a factor, the remainder will equal 0.
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Let's go ahead and do that: here we are dividing by 2, and I do not have any missing terms.
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I have a coefficient of 1, a coefficient of -3, a coefficient of 2, another coefficient of 1, and a coefficient of -8.
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Bring down 1; multiply 1 by 2; that gives me 2.
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And now, I take 2 plus -3 to get -1; multiply -1 times 2 to get -2.
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Combining 2 and -2, I get 0; 0 times 2 is 0; 1 + 0 is 1; 1 times 2 is 2, to give me a final value of -6; this is the remainder.
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2 is not a zero of f(x); if the remainder here was 0, then this would tell me that 2 is a zero of f(x),
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because it is telling me that f(2) equals 0; in other words, when x = 2, the function equals 0.
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And by definition, that is a zero; it is crossing the x-intercept.
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OK, to sum up: if you are looking for rational zeroes, list all possible rational zeroes using the theorem.
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And the theorem says that p/q equals the factors of the constant, over the factors of the leading coefficient.
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Once you have done that, use synthetic division to determine which possibilities are actually zeroes.
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Division will yield a remainder equal to 0, if the value is a zero.
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So, you get your values for p and q; you take the function, and you divide it by each of the values that you are trying, p/q.
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And if the remainder equals 0, p/q is a zero.
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If the remainder is anything other than 0, then it is not.
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So, list the rational zeroes, and then use synthetic division to check.
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OK, list all possible rational zeroes of this function.
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I need to find values of p/q, and I am going to first look at the constant; this is a₀ = 9, and the factors are 1, 3, and 9.
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So, these are values for p.
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Here, the leading coefficient equals 2; factors are just 1 and 2.
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So now, I am going to go through all of these different combinations of numerator (1, 3, 9) and denominator (1 and 2).
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So, possible rational zeroes are ±1/1 (I am just going to simplify that to 1 right away), or ±1/2.
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Next is ±3/1 (which is just 3); ±3/2; next, ±9/1 (which gives me 9); ±9/2.
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So, I have 1, 2, 3 times 2; that is 6 possibilities; and then plus or minus for each gives me 12 possible rational zeroes.
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I don't know which, if any, of these are actually zeroes; I would have to check using synthetic division.
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In this next example, we are asked to actually find the rational zeroes, not just to list the possibilities.
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So, I am going to need to use synthetic division.
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Since the leading coefficient is 1, then possible rational zeroes equal the factors of 3.
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I don't need to worry about this; the denominator is just going to be 1.
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OK, factors of 3 are 1 and 3; so my possibilities are ±1 and ±3.
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Now, I need to check these using synthetic division.
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So, I am starting out with 1; and I am looking, and there are no missing terms: x³, x², x, and a constant.
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So, I can just go ahead and put these coefficients: 1, -2, -2, -2, -3.
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This is 1 times 1 is 1; combining that with -2 gets -1; 1 times -1 is -1, plus -2 is -3; -3 times 1 is -3; this is -6.
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The remainder equals -6, so 1 is not a zero.
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OK, now let's try -1 right here: again, my coefficients are the same: 1, -2, -2, -3.
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OK, this is going to give me 1 times -1 (is -1); -2 and -1 is -3; -3 times -1 is 3.
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I am combining that with -2 to get 1; times -1 gives me -1 and -3; this is -4; so this is not a zero.
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OK, the next possibility: let's try 3; the coefficients are 1, -2, -2, -3.
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This is going to give me 1; times 3 is 3; 3 and -2 is going to give me 1.
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1 times 3 is going to give me 3 again; and -2 is going to give me 1.
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1 times 3 is 3; 0; 3 is a zero, because the remainder equals 0; 3 is a zero.
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OK, I found one rational zero: I have one more possibility left--I checked 1; I checked -1; I checked 3; now I need to check -3.
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1, -2, -2, -3: bring down the 1 and multiply by -3 to get -3.
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Combine with -2 to give me -5; -5 times -3 is 15; plus -2 leaves me with 13; times -3 is -39.
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And that is -42, so this is not a zero.
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The rational zero for this function is 3; 3 is a rational zero.
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I had this list of possibilities, and the only one that turned out to be correct, an actual zero, is 3.
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List all possible rational zeroes of this function.
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So, I am going to be looking for p/q, which equals the factors of the constant over factors of the leading coefficient.
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What this is going to give me is 09 1, 3, 5, and 15, over factors of 4: 1, 2, and 4.
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So, I need to figure out all of these possibilities and use positive and negative for each.
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So, 1/1 is ±1; and then, 1/2 is ±1/2; 1/4 is ±1/4.
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OK, now going on to 3: 3/1 is just ±3; 3/2 is ±3/2; 3/4 is ±3/4.
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OK, next, 5: 5/1 is just 5; 5/2; 5/4...
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Then, 09 15/1--just 15; 15/2; and then 15/4, plus or minus.
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So, there are a lot of possibilities here; and I listed them all out.
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But if you were actually asked to find the rational zeroes, you would need to check these through synthetic division.
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OK, here we are asked to find all zeroes of this function; and notice, it doesn't just say "rational zeroes"; it says "all zeroes."
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But I am going to start out by looking at the rational zeroes.
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And since the leading coefficient is 1, the possible rational zeroes will equal factors of the constant.
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And that is because, if I am taking p/q, and the denominator is the factors of the leading coefficient...
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here the only factor is 1; so I am just going to end up with p.
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So, factors of 4 are ±1, ±2, and ±4.
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Now, I have a list of possible rational zeroes, and I am going to use synthetic division to determine which of these are actual zeroes (not just possibilities).
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The coefficient here is 1; the coefficient here is 2; -2; and -4.
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Now, there are no missing terms: I have cubed, squared, x, and constant.
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So, I am going to go ahead and use synthetic division; this is 1 times 1, is 1.
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So, this gives me 2 and 1, which is 3; 3 times 1 gives me 3; that combined with -2 is 1.
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1 times 1 is 1, combined with -4 is -3.
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This is not a zero: 1 is not a zero, because the remainder is something other than 0.
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Let's try -1: the coefficients are, again, 1, 2, -2, and -4.
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Here, I have 1 times -1 to yield -1; combined with 2 is going to give me 1; this times -1 is -1.
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Combined with -2, it is -3; -3 times -1 is 3; plus -4 is -1; again, this is not a zero.
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So, not a zero, not a zero...let's try 2.
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OK, again, same coefficients, same process: bring down the 1; multiply by 2; combine to get 4.
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4 times 2 is 8; plus -2 leaves me with 6; times 2 is 12; plus -4 is 8--again, not a zero.
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Now, I am going to try -2: bring down the 1; times -2; combining 2 and -2 leaves me 0, times -2...
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Combining, you get -2 times -2, is positive 4; -4 and positive 4 is 0.
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So, looking at this, since the remainder equals 0, -2 is a zero of this function.
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Now, I could continue on to check these two; but there is an easier way to proceed.
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because what this is telling me is that (remember, this is the opposite sign) x + 2 (because I took the opposite sign) is a factor of this.
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So, that leaves me with x + 2, times...this is x³, so times 1x², plus 0x (that drops out), minus 2.
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This is what I am actually left with: (x + 2) (x² - 2).
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So, instead of just checking these two (and they might be right; they might not be right),
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what I am going to need to do, if these are not correct (because there are three zeroes, because the degree is 3)
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is: then I am going to need to find irrational or complex zeroes.
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So, instead of bothering to even check these, I just use the fact that I now have (x² - 2) (x + 2) to go ahead and use factoring to solve.
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So, I know that I have x + 2, and I look, and I can't factor any more.
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So, I use the zero product property; and I know already that x + 2 = 0, and x = -2; I figured that out.
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But I can go on and do it with this, as well: x - 2 = 0.
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And this is going to give me x² = 2; I am then going to take the square root of both sides to get x = ±√2.
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I have two zeroes here, and I have one here: the zeroes are -2, √2, and -√2.
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And this is an irrational number, so it wasn't on my list of possibilities.
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It is easier to proceed by finding one zero and then using the information that you have here to write the factored-out
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(or at least partially-factored-out, in some cases) form of the equation,
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and then go on and use the zero product property to find the remaining zeroes.
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Thanks for visiting Educator.com, and I will see you next lesson!