>> Here we have a third degree polynomial function, we'd like to identify the three linear factors and also the three 0s associated with this function. We would begin the process by perhaps thinking what is F of 0; well when X is replaced with 0, these terms go out and we have negative 10. Now F of 1 is pretty easy to identify just visually, if X is replaced with 1, we would 1 minus 6 plus 13 minus 10, collecting we find that value to be negative 2. So 1 goes to negative 2, then we would look at this little table and we'd think well, let's see, 0 is at negative 10 and 1 is at negative 2, the graph is sort of approaching the X axis, maybe we get some mileage by investigating at 2. Well let's see, investigating synthetically, 2 into now the 1, the negative 6, the 13 and the negative 10. Let's see, bring down the 1, 2 times 1, 2; together here, negative 4. 2 times negative 4, negative 8; together, 5. 2 times 5, 10 and with negative 10 gives 0. So we do indeed have a 0 at 2 so we have a 0 at 2 and X minus 2 then is a factor so we can go through and factor. [ Background noise ] And the leftover factor we just read down here, this is 1X squared minus 4X then plus 5. And now we would look to factor here, well we would investigate as we will but we find out that we can't factor using the methods that we've been talking about and so we need to find the 0s by using the quadratic formula. So using an A value of 1, B of negative 4 and a C of 5, we come over to the side and we apply the quadratic formula. Write the formula down, replace within it in the next step now notice that B is negative 4 so this is minus negative 4. You see, the opposite of negative 4 plus or minus B squared is negative 4 squared minus 4 times A is 1, C is 5. In the denominator, 2A means 2 times 1. Now let's begin to evaluate; minus negative 4 means 4 plus or minus. Now under the radical, we get negative 4 when the smoke clears on that business. Now the square root of a negative number, well we have to be a little cautious here, we're talking about an imaginary value here so our answers are going to be purely complex. All right, let's go through the process though, the square root of negative 4 is -- let's see, the square root of 4 is 2 and the square root of the negative 1 is I so we have plus or minus 2I. Now 2 divides both parts of this numerator so we can simplify to 2 plus or minus I; now we've identified all of our 0s, our 0s are 2 plus I, 2 minus I and 2. You see, those are our 0s but let's go through the factored -- the factorization, now the 2 plus or minus I, 0s give us factors as X minus each of those so if we want to write a factored form, then we're writing it this way. Now for the other 0s, it is X minus 2 plus I, X minus 2 minus I, kind of getting jammed with this other problem here. Let me erase this. So that's the idea and now we just simplify, we just clean up a little bit in those brackets. You see and we can just drop the parenthesis, you know, change the signs, drop the parenthesis here so this is X minus 2 minus I and here is X minus 2 plus I as our three linear factors and notice that the complex 0s, the 2 plus or minus I will kind of naturally in the flow of things here, appear in conjugate pairs and that is universally true that complex 0s occur always in conjugate pairs.