Finding zeros of polynomials (2 of 2)

- [Voiceover] In the last video, we factored this polynomial in order to find the real roots. We factored it by grouping, which essentially doing the distributive property in reverse twice. And I mentioned that there's two ways you could do it. You could actually, from the get go, add these two middle degree terms, and then, think about it from there. So, what I thought I'd do is just a quick video on that alternative. So, if we add, instead of grouping, if we add these middle two terms. Actually, I'll just focus on the fourth degree polynomial here. We know that we have an x out front. This fourth degree polynomial is going to simplify to x to the fourth plus seven x squared minus 18. If we wanna factor this, we could recognize a pattern here. You probably remember. Hopefully you remember. If you don't, then you might wanna review your factoring polynomials. But if you have x plus a times x plus b, that's going to be equal to, x is going to be equal to x squared plus the sum of those two numbers, a and b, as being the coefficient of the x term plus the product of those two numbers. If you just multiply this out, this is what you would get. But if this was x squared plus a times x squared plus b, instead of this being x squared, this would be x to the fourth. Instead of this being x, this would be x squared, which is exactly the pattern we have here. So, what two a's and b's that if I add them up, I would get seven, and if I were to take their product, I get negative 18? Well, since their product is negative, we know that they are of different signs. One will be positive, one will be negative. And since their sum is positive, we know that the larger of the two numbers is going to be positive. So, what jumps out at me is nine times negative two. You multiply those, you get negative 18. You take their sum, you get seven. So, we can rewrite this, just looking at this pattern here as x squared plus nine times x squared minus two. I could say plus negative two. That's the same thing as x squared minus two. And then, that's exactly what we got right over here. Of course, you have this x out front that I didn't consider right over here. And then, this, as we did in the previous video, you could recognize as a difference of squares and then, factor it further to actually find the roots. But I just wanted to show that you could solve this by regrouping, or you can solve this by, I guess you could say, more traditional factoring means. And notice this nine and negative two, this is what was already broken up for us, so we could factor by regrouping.