WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we will be working on solving polynomial equations.
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And we are going to start out by reviewing some factoring techniques.
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And the techniques that you will use for factoring polynomials are familiar from earlier work.
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And these include greatest common factor, difference of two squares, perfect square trinomials, general trinomials, and factoring by grouping.
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So, just as we did with quadratic equations, if you are going to be working with a polynomial,
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the first thing you want to do is: if there is a greatest common factor, factor that out first.
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For example, if I had something like 4x⁴ - 6x² - 12x, I have a greatest common factor of 2x.
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So, I have pulled that out first; and that would give me 2x, leaving behind 2x³; here that would leave behind 3x; and here, -6.
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And then, I would work on factoring that farther, if it is possible.
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OK, the difference of two squares: this is the greatest common factor, and now difference of two squares.
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Recall that these are in the form a² - b², and an example would be x² - 9.
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This could be factored out into (x + 3) (x - 3); this factors into (a - b) times (a + b).
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Here, in this case, a equaled x and b equaled 3: x² - 3².
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In a few minutes, we will be talking about the difference and sum of two cubes; we will go on a little farther than just working with squares.
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Next, perfect square trinomials: you know that, if you recognize these, they are easy to work with.
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So, you want to be on the lookout for these perfect square trinomials.
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For example, one that we have seen earlier, working with quadratic equations, is x² + 8x + 16.
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And this factors as (x + 4)², or (x + 4) (x + 4)--a perfect square trinomial.
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And all of these are covered in detail in earlier lectures, so go ahead back and review these, if necessary,
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to make sure you have them down before you work on factoring polynomials.
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General trinomials: you would recognize these in a form that they don't fit into the special cases
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of difference of two squares or perfect square trinomials.
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It might be something such as, say, x² + 2x -8; and you need to just use some trial and error on this.
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For example, for this one, you would say, "OK, the first term in each factor has to be x."
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And since I have a negative in front of the constant here, I have to have a positive here and a negative here,
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because a positive times a negative is going to give me a negative.
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Then, I am going to look at the factors of 8, and I am going to say, "OK, I have 1 and 8, and 2 and 4."
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And then, I want to find factors of 8 that, when one is positive and one is negative, sum up to 2.
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And I can see that 1 and 8 are too far apart; they are never going to give me 2.
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2 and 4 are perfect; and I see that, if I make the 4 positive and the 2 negative,
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I am going to get 2, which is going to be the coefficient for my middle term.
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This is going to factor out to (x + 4) (x - 2); and I can always check this by using FOIL,
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which will tell me that the First term gives me x²; Outer is -2x; Inner is 4x; and then, this is -8.
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So, x² + 2x - 8--it gave me this trinomial back.
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Grouping: recall that we use grouping when we are factoring polynomials with 4 terms--factoring by grouping.
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Let's say you have some things such as 3x³ - 4x² + 6x - 8.
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I am going to handle this by grouping the first two terms, and then grouping the second two terms.
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Now, I am going to pull out any common factors I see; and I do have a common factor of x², to leave behind 3x - 1.
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Over here, I have a common factor of 2; I am going to pull that out, and that is going to leave me with 3x - 4.
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I am pulling out the x²: that is going to leave me with 3x - 4 here; here I am going to pull out the 2; that is going to leave me with 3x - 4.
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This is what I wanted to happen; and what I have here is a common binomial factor.
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I am going to factor that out (pull that out in front), and that leaves behind x² + 2.
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So, factoring by grouping: group the first two terms; group the second two; pull out a common factor.
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If you have a common binomial factor left behind, then you pull that out in front.
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Make sure you are familiar with all of these; review the ones that you need to.
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And we are going to be using these to solve polynomial equations.
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Now, we are going on to a new concept with factoring: and that is the sum and difference of two cubes.
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You are familiar with working with squares; and now we are talking about the sum and difference of two cubes.
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Let's look at an example: 64x³ + 125.
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Well, if I think about this, the cube root of 64 is 4; the cube root of x³ is x; the cube root of 125 is 5.
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Here, this is in this form, a³ + b³; here, a equals 4x, and b equals 5.
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Now, by memorizing this formula, I can just factor this out.
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So, if I know that this is in this form, a³ + b³, which equals
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(a + b) times (a² - ab + b²), then I know that I have that a is 4x and b is 5.
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So, I just have to substitute that in.
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Here, I am supposed to have a², so that is going to give me (4x)², minus a times b, plus b².
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OK, this is 4x + 5, and this is going to give me 16x² - 20x + (5² is) 25.
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So again, just recognize that this is actually the sum of two cubes, so it is going to factor out to this form.
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For the difference of two cubes, it is a very similar idea--only you are going to end up with a negative sign here and a positive here.
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It is the same idea, just different signs.
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So, 64x³ + 125 factors out to this.
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And then, from there, in some cases, you may be able to go on with your factoring.
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Let's talk about quadratic form: we worked with quadratic equations earlier on, and you can actually
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put higher-order polynomials into a form called **quadratic form**.
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And quadratic form is an² + bn + c, where n is an expression of x.
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Well, what does that mean--"n is an expression of x?"
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Let's use an example: let's say I have 6x⁸ + 5x⁴ + 9, and I want to get it into this form, an² + bn + c.
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Well, let's look at this more closely: I want to get this into some form where I have a number squared, so I have a square here.
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Now, I am going to think about: "x to the what, squared, equals x⁸?"
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Thinking about raising a power to a power, I know that I have to multiply 2 times something to get 8.
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So, 8 divided by 2 is 4; so that tells me that (x⁴)² is x⁸.
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So, all I did is took my x⁸ and put it in a form where it is something squared.
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Now, I am going to assign another variable this value, x⁴.
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I could use y; I could use z; I am going to go ahead and use y--I am going to let y equal x⁴.
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Therefore, y² equals x⁸, since y equals x⁴,
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because if y equals x⁴, I could just substitute and say that this is really (x⁴)².
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OK, so again, what I did is looked at what I had, x⁸; and I said,
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"What would have to be the exponent here, so that I could just square it and get this back?"
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Well, the exponent would have to be x⁴.
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Then, I took this and wrote it as a different variable, y = x⁴.
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I assigned the variable y the value x⁴; so that is what this means--something n².
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Here, I am just going to say it is y; so this gives me...my leading coefficient is 6.
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Now, instead of putting n, I just decided I am going to use y to represent x⁴, and that is going to be y².
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Plus...my next coefficient is 5.
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OK, here I have an x⁴, but I said y = x⁴, so I am just going to write it as y.
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And then, my constant...so now, I have taken this and rewritten it as this equation, where y is equal to x⁴.
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Now that this is in quadratic form, I can use techniques such as the quadratic formula and factoring, learned earlier on,
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in order to (if this were an equation, let's say, then I could) solve it using these techniques for working with quadratics.
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Once I find the value of y, whatever y turns out to be, if I know y, I can find x.
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If y is 16, then I would just have to say, "OK, this equals 2"; then x would equal 2.
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So, by putting this equation into quadratic form, I can use quadratic techniques to solve for y.
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Once I have y, I can solve for x, which makes quadratic form very useful when working with certain polynomials.
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Looking at this first example, 27a³ - 8, I recognize this as the difference of two cubes.
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And this is in this form; here, the cube root of 27 is 3; the cube root of a³ is a; so this is what I have.
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The cube root of 8 is 2; therefore, what I have in here is a, and what I have right here is b.
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And sometimes it helps to actually write these out instead of doing it in your head, just to keep everything straight.
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Since I know that a is 3a, and b equals 2, I just have to substitute here to factor: a - b--I am going to rewrite it down here:
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a² + ab + b²; OK, if a is 3a, then that is going to give me 3a - b (which is 2).
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Here, I am going to have a²; so that is 3a² + a times b + b².
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This is going to give me 9a² + 6a + 4.
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And one important thing is just to watch the signs, working with the difference of two cubes or the sum of two cubes.
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And then, figure out what your a and b values are by finding the cube root of these terms.
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And then, just substitute in a and b into this formula to get factorization.
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In the next example, we are asked to factor this trinomial; and the first step is to factor out a greatest common factor, if there is one.
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And I am looking here for a greatest common factor.
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And I see that I have a 3 that can be pulled out of all of these terms.
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For x's, I have x²; I have an x here; and an x³; so they all have at least an x.
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For y, there is one y here; there is a y³ here; and there is a y² here.
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For z, I have z³, z², and z⁴, so they all have at least z².
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So, the greatest common factor is 3xyz²; so I am going to pull that out and see what is left behind.
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In my first term, the 3 is gone; I am left with an x.
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The y is gone, and the z² is gone; that leaves me just xz.
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Here, I have a negative sign; I pulled a 3 out, leaving behind a 4; the x is gone; one y is gone, so I have y²; z² is gone.
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Here, I pulled out the factor of 3; that leaves behind 8; one x is gone; I have x²; one y is gone; and z² is gone.
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OK, so I am left with the greatest common factor, times what was left behind in here.
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And I am looking, and there are no common factors left.
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I don't recognize this as the difference of two squares, or anything that I can factor out.
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So, this is as far as I can go with the factorization.
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Now, I am asked to write this polynomial in quadratic form; recall that quadratic form is an² + bn + c.
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Here, I have x⁶; I want to find some variable, assign it a certain value, and then square it.
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So, x to the something, squared, equals x⁶.
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Well, if I take 6 and divide it by 2, I am going to get 3; therefore, (x³)² = x⁶, because 3 times 2 is 6.
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I am going to let y equal x³; therefore, y² is going to equal x⁶, because that would be (x³)².
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Now, let's look at what we have: we have our leading coefficient of 10, x⁶...
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well, x⁶ is going to be the same as y²; minus 12; now, I have x³ here.
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Well, I said that y equals x³, so I am just going to put y here; plus 14.
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This is my given equation (this polynomial) in quadratic form; and again, I did that by assigning y the value of x³,
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which allowed me to write that as 10y² - 12y + 14.
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I could then use methods for solving quadratic equations, if this were an equation.
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Then, I could solve this and find y; once I have y, I could go back in and just find the cube root of that, which would give me x.
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In this example, I am actually asked to solve the equation; I am going to start out by factoring.
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And this is a trinomial; and again, when you see that x⁴ trinomial, you could think
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that it is similar to working with a quadratic, except you have x² here in the factors, instead of just x.
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So, x² times x² is x⁴.
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I have a negative sign here, which means that one term is going to be positive, and one is going to be negative (for the second terms).
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Let's think about the factors of 24, and let's find the factors that are going to add up to -2.
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Factors of 24: 1 and 24, 2 and 12, 3 and 8, 4 and 6.
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Now, this is a lot, but I don't have to work with all of these; I want to find ones that are close together, because their sum is just going to be -2.
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So, the closest together are 4 and 6; and I see that one is positive and one is negative.
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Well, if I make 4 positive and 6 negative, their sum will be -2.
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So, I know that this is what I want: I want for 4 to be positive, and I want the 6 to be negative (and this all equals 0).
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OK, so I have actually factored this as far out as I can; x² + 4 doesn't factor any farther, nor does this.
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So, I am just going to go ahead and use the zero product property to solve.
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And the zero product property tells me that x² + 4 = 0, or x² - 6 = 0,
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because if either of these terms is 0, the product of the two will be 0, and the equation will be solved.
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Here, I am going to get x² = -4.
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Now, when I take the square root of that, I am going to end up with this.
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Recall from working with imaginary and complex numbers that the square root of -1 equals i.
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So, this is the same as saying, "The square root of -1 times 4," which equals the square root of -1, times the square root of 4.
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So, since this equals i, I am just going to write this as this; and that is actually plus or minus;
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we have to take the positive and the negative, according to the square root property.
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OK, well, this is a perfect square, so this gives me ± 2i--that is some review of complex numbers.
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Here, it is a little bit simpler: x² = 6.
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According to the square root property, if I take the square root of both sides, then I will get x = ±√6.
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All right, so I was asked to solve this; and I came up with solutions.
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And the solutions are that x equals 2i, -2i, √6, or -√6.
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And I figured that out by factoring this into these two factors, (x² + 4) times (x² - 6),
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and then continuing on to use the zero product property.
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In this case, I ended up with x² equaling a negative number, so I ended up having to use a complex number for my result.
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And then, these are the four solutions to this polynomial equation.
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That concludes this session of Educator.com on polynomial equations.
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And I will see you next lesson for more work with polynomials.