9-12 N.CN.8 Transcript
This is Common Core State Standards Support Video for Mathematics. This is a high school standard for grades 9-12 N.CN.8.
This standard states: Extend polynomial identities to the complex numbers. This is pretty short and simple. The focus is applying the polynomial identities so let's look at those. One identity is a product of two binomials and these are pretty common, (x + a) times (x + b), would come out to be x squared + (a + b), the sum of a and b times x + a b as our constant, then (x – a) times (x – b). The signs are both negative, and we get x squared - (a + b) times x + a b for our constant. Then we have the situations where the two binomials have different signs, so we have (x – a) times (x + b). Now in this scenario, if the absolute value of a is more than the absolute value of b the result will be x squared minus the difference of the absolute value of a and b as our coefficient of x minus the product of a and b as our constant. Then in the scenario where the absolute value of a is less than b, we would get x squared plus the difference of the absolute value of a and b times x minus ab. We also have our binomial squares (a + b) all squared to be a squared plus 2ab plus b squared. We have (a -b) all squared would be a squared minus 2ab plus b squared. Very common identity the difference of two squares, a squared minus b squared, would factor out to be (a - b) times (a + b). Then we have our quadratic formula for use in those cases where we have a quadratic that does not factor. Given the general case, ax squared plus bx plus c = 0, then our solutions for x would be the opposite of b plus the square root of b squared minus 4ac all over 2a. Our second solution would be the opposite of b minus the square root of b squared minus 4ac all over 2a. Then we also have polynomial identities for cubes. If we have two cubes, a cubed plus b cubed, that would factor out to be (a + b) times (a squared - ab + b squared). Then we have the difference of two cubes, a cubed minus b cubed, that would be (a -b) times (a squared + ab + b squared).
So let's take the first identity, product of two binomials. How does that relate to the complex numbers? Well the closest parallel will be if we take the product of two complex numbers, so we take one complex number (a + bi) with a being the real part and the i being the imaginary, times a second complex number (c + di). If we multiply those out, use our distributive property we notice that for the two middle terms we have i as a common factor. So we factor that out, and then we can simplify the i squared to - 1. Now we switch the order of our - 1 to have just - bd for our last term. Now we change our order around by using our commutative and associative properties to put this in the correct format for representing a complex number. So (ac - bd) would be the real part of our complex number, and (bc + ad) times i would be the imaginary component. Now notice the patterns for the product of two binomials. We always have the same first term in our two binomials but in looking at the product of our two complex numbers our first two variables and in this case for the complex numbers it was the real parts they’re not the same. So we can't apply the pattern of the products of the two binomials to the product of two complex numbers. The best we can do is, that the only thing they have in common is just that, it’s a general application of the distributive property in both cases. What about a binomial squared? Well if we take the complex numbers and look at a complex number squared, we start off with (a + bi) as our complex number and square it, multiplied by itself. So we use our distributive property, do our multiplication, combine our middle terms, change our i squared to - 1, do some simplification, so we have a squared plus 2abi minus b squared. Then we change the order around to fit the representation for a complex number. So we get (a squared - b squared) as our real part and 2abi as our imaginary. The complex numbers that we used initially was the sum of the real and the imaginary parts. If we were to take complex numbers where it's the difference of the real and the imaginary parts and take that number and square it, we get a similar result, the difference being the sign of our middle term. If we look back at our real number situation with the binomial squared notice that we do have some similarities. We have the 2ab, the middle term for our binomial squared does turn out to be the coefficient for our i in both cases. Then when we put this in complex number format, notice that with our binomial squared we had plus b squared, but with our complex number because of changing the i squared to negative one we get minus b squared instead. If we were to take some complex numbers and square them, let’s take this example here, on the left (3 + 2i) times itself if we take this general pattern and apply it, well we did it the long way but let's see what would happen if we apply this. This says that I'd take the difference of the squares. So if I take the a and our b as 3 and 2, that would be 3 squared is 9, 2 squared is 4, 9 - 4, it does turn out to be 5. Then this says take the product of a and b and double it. So we would take 3 x 2 = 6, double it that is 12. So 12i would be our imaginary part, so that checks out. Then for our second example over here we take (4 - 5i) and multiply it by itself. According to this, we take the difference of the squares so that would be 16 – 25, that is a - 9. For our imaginary part we take a x b which would be 4 x 5 = 20, double it, that is 40, so it is - 40i, so it checks out. So we can apply the binomial squared identity to the complex numbers to make our job a little bit easier.
What about the difference of two squares? So applied to a complex number, if we multiply (a – bi) times (a + bi) we would get a squared - abi + abi - b squared i squared, and our two middle terms add up to be zero. Then we can change our i squared to -1 which results in a squared + b squared. Now so instead of having the difference of two squares we have the sum of two squares for our complex number. So let's look at an example of applying the difference of two squares to complex numbers. We already saw what happens when we take (a – bi) times (a + bi). We got a squared plus b squared. So if we take for example (2 - 7i) times (2+7i) and we do all of the work we end up with 4 + 49, which will give us 53. But if you go directly from what we get from the multiplication a squared plus b squared, this says we could have just taken our a term which in this case was two squared and we get four and our b term which is a 7 multiply that, that's square, that that's 49. We could have gone directly straight to here. Let's take another situation with the difference of two squares as applied to complex numbers. Let's say we have x squared plus 49 is equal to 0. Now we can tell by looking that x squared that's going to give us a positive number and then we are going to add it to 49. So we know that there's no way that we can combine these two positive numbers to be zero, so we know our answer is going to be complex. So if we take our pattern for the multiplication of two complex numbers (a – bi) times (a + bi), we take our situation x squared plus 49 is equal to 0 and directly apply our factoring. So we can solve this then separately, we know that this first one here would come out to be a + 7i for our solution. This one would come out to be - 7i for our solution. Of course an alternative would be to solve it this way, just take our original equation and subtract 49 from both sides. So x squared is equal to - 49 take the square root of both sides and so we get + or - the square root of - 49 which of course will be + or - 7i.
What about the quadratic formula applied to complex numbers? Let's put this in the general form the way that we're used to seeing it with the plus and the minus here. Now we know that the (b squared - 4 ac) part is called the discriminant and that's really the concern here. If the discriminant comes out to be positive, we're going to get two real roots. If the discriminant is zero we're going to get one double root. If the discriminant is negative, we're going to get two complex roots. So really this is our concern here, our application to the complex numbers for the quadratic formula. That's going to be the situation if the discriminant is negative because we know we're going to get complex solutions. So let's take an example 3 x squared + 2 x plus 5 = 0, apply our quadratic formula, and substitute the numbers. Let's do the discriminant. First we end up with 4 - 60, which is a - 56. So we automatically know already that we're going to get two complex roots. So we finish our computation - 56 and that factors out to a - 4 x 14. Take out the square root of - 4 as 2i + or -. Do a little bit of simplification, so there's our final solution.
Let's look at cubes the sum of 2 cubes and the difference of 2 cubes applied to complex numbers. Well let's take a situation where we have let's say x cubed + 64. Let's change that to x cubed plus 4 cubed so we can apply our identity. So if we follow our pattern for our identities our x cubed + 4 cubed would be (x + 4) times x squared - 4 the product of 4 and x + 4 squared. Do a little bit of simplification and here's where we are, we have (x + 4). So we know that's going to be a real solution, but what about the (x squared - 4x + 16). Well typically with real numbers, we would just solve it this way. Just subtract 64 and the cube root of - 64 is - 4. So that takes care of that, but we're going to have three solutions. So now we have to worry about the x squared minus 4 x plus 16 is equal to 0, like we started to a while ago. So we need to be concerned about that so what solutions would we get here? Well let's take our x squared - 4 x plus 16 i = 0 and apply our quadratic formula. Our concern initially is the discriminant, what type of solution will we get. So let's apply the discriminant here and substitute our values 16 - 64 and we get a - 48. So automatically we know we're going to get for our other two roots a complex number. Ok, we get - 48 so we have a negative for our discriminant and we have a negative square root. So we know since the discriminant is negative, we're going to get two complex solutions. Now with x cubed + 64, we got one real solution and two complex ones. Now is that going to be typical, is that going to be the general case? So let's look and see if we can figure that out. So let's look and do a little bit of investigation and see if that is a valid conclusion. We've already figured out that this first part the (x + b), that's going to give us our one real solution. So let's look at the trinomial here and see what happens with the discriminant, when we apply the quadratic formula. So we take our general form for the quadratic formula and let's just worry about the discriminant. We plug in our values for b, a, and c, solve, and we get b squared minus four b squared, which of course simplifies to - 3b squared. Now we know that b squared is going to give us a positive number times a - 3 will result in the negative number. So the discriminant will be negative resulting in two complex solutions.
If we take this a step further and consider possible situations where for our x cubed we have a constant that’s also a perfect cube. What happens there? Well if we do the exact same thing, take our trinomial and apply that to our quadratic formula but we're only going to worry about the discriminant. We plug in our values for a c and b and here they are. So when we simplify we get - 3b squared d squared and again we know that b squared and d squared are going to give us positive numbers times a - 3. So again the discriminant will be negative. So yes we will get two complex solutions. Now that was the sum of two cubes what about the difference of two cubes? Does that apply here also? So let's go straight to the general case where we have a perfect cube also for a coefficient for our x cubed term. Is it the same situation where we'll get one real root and two complex ones? Here's our concern again, the trinomial, and we're going to do exactly what we did before. We're going to apply this trinomial and use our quadratic formula. We're only worried about the discriminant, substitute our values just like we did before, solve, and again we get -3b squared d squared. Again we know that this is going to give us a negative number because b squared and d squared will give us positive numbers times a - 3. So yes the discriminant will be negative again resulting in two complex solutions. You might consider taking this a step further but it really is beyond the scope of this standard. But you can look at the general graph of the sum or difference of two cubes and you'll see that you'll always have just one x intercept when you set it equal to zero. So we will get just one real solution making the other two complex. So this standard again: extend polynomial identities to the complex numbers. Not as simple as it looked because there's quite a few scenarios with the polynomial identities that we had to consider.