ਬਹੁਪਦਾਂ ਦੇ ਸਿਫ਼ਰ ਅਤੇ ਉਹਨਾਂ ਦੇ ਗ੍ਰਾਫ਼

Use the real 0's of the polynomial function y equal to x to the third plus 3x squared plus x plus 3 to determine which of the following could be its graph. So there's several ways of trying to approach it. One, we could just look at what the 0's of these graphs are or what they appear to be and then see if this function is actually 0 when x is equal to that. So for example, in graph A-- and first of all, as always, I encourage you to pause this video and try it before I show you how to solve it. So I'm assuming you've given a go at it. So let's look at this first graph here. Its 0, it clearly has a 0 right at this point. And just by trying to inspect this graph, it looks like this is at x is equal to negative 3, if I were to estimate. So that looks like the point negative 3, 0. So let's see, if we substitute x equals negative 3 here, whether we get y equaling 0. So let's see, negative 3 to the third power plus 3 times negative 3 squared plus negative 3 plus 3. What does this give us? This gives us negative 27. This gives us positive 27. This gives, of course, negative 3. This is plus 3. These two cancel out. These two cancel out. This does indeed equal 0. So this was actually pretty straightforward. Graph A does indeed work. You could try graph B right here, and you would have to verify that we have a 0 at, this looks like negative 2. Another one, this looks like at 1, another one that looks at 3. And since we already know that A is the answer, none of these-- if you were to input x equals negative 2, x equals 1, or x equals 3 into this function definition right over here, you should not get 0. And you'll see that this doesn't work. Same thing for this one. If you tried 4 or 7 for your x's, you should not get 0 over here, because we see that the real function does not equal 0 at 4 or 7. Another giveaway that this is not going to be the function is that you are going to have a total of three roots. Let me write this down. So you're going to have a total of three roots. Now, , those three roots could be real or complex roots. And the big key is complex roots come in pairs. So you might have a situation with three real roots. And this is an example with three real roots, although we know this actually isn't the function right over here. Or if you have one complex root, you're going to have another complex root. So if you have any complex roots, the next possibility is one real and two complex roots. And this right over here has two real roots. That's not a possibility. That would somehow imply that you have only one complex root, which that is not a possibility. Now another way that you could have thought about this-- and this would have been the longer way. But let's say you didn't have the graphs here for you, and someone asked you to just find the roots-- well, you could have attempted to factor this. And this one actually is factorable. y is equal to x to the third plus 3x squared plus x plus 3. As mentioned in previous videos, factoring things of a degree higher than 2, there is something of an art to it. But oftentimes, if someone expects you to, you might be able to group things in interesting ways, especially when you see that several terms have some common factors. So for example, these first two terms right over here have the common factor x squared. So if you were to factor that out, you would get x squared times x plus 3, which is neat because that looks a lot like the second two terms. We could write that as plus 1 times x plus 3. And then you can factor the x plus 3 out. We could factor the x plus 3 out, and we would get x plus 3 times x squared plus 1. And now, your 0's are going to happen, or this whole y-- remember this is equal to y-- y is going to equal 0 if either one of these factors is equal to 0. So when does x plus 3 equal 0? Well, subtract 3 from both sides. That happens when x is equal to negative 3. When does x squared plus 1 equal 0, I should say? Well, when x squared is equal to negative 1. Well, there's no real x's, no real valued x's. There's no real number x's such that x squared is equal to negative 1. x is going to be an imaginary-- or I guess I'll just say it in more general terms-- it's going to be complex valued. So once again, you see you're going to have a pair of complex roots, and you have one real root at x is equal to negative 3.