Need to figure out two numbers that add up to the coefficient See a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say alright well I Now, I did it fairly involved mainly so you see where all this came from. We've just factored it asĪ product of two binomials. I could just write x minusįive, and we're done. Negative five right over there because that's our b. Of saying plus negative five which we could say, we could just say, actually let me write that down. Is going to be equal to x plus two times x, instead And so our original expression, we can rewrite as, so we can rewrite x squared We could say that b is two, but I'll just say that a is equal to two and b is equal to negative five. Two plus negative five is going to be negative three. Now this is interesting because still when you multiply them But if you add these two, you're going to get positive three. Talking about negative 10, you could say negative two times five. And two and five are interesting because if one of them are negative, their difference is three. So if I were to just factor 10, 10 you could do that as one times ten, or two times five. We get a negative number, we know that the negative And so let's see how weĬould think about it. Them we get a negative number, we know that they're going Well what could those numbers be? Well since when you multiply Up to negative three, to add up to the coefficient here. ![]() And if I multiply those same two numbers, I'm going to get negative 10. On the first degree term, so two numbers that add A way to factor it is toĬome up with two numbers that add up to the coefficient On the x squared, you don't even see it but You're factoring something, a quadratic expression that hasĪ one on second degree term, so it has a one coefficient Well a times b needs toīe equal to negative 10. So we have a plus b needs toīe equal to negative three. They need to add up toīe this coefficient. So one way to think about it is that a plus b needs to beĮqual to negative three. ![]() We have something times x, in this case it's a If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there. And now we can use this to think about what a and b need to be. a plus b x's, and thenįinally I have the plus, I'll do that blue color,įinally I have it. ![]() If I have ax's and I add bx's to that I'm going to have a plus b x's. The first degree terms, they're both multiplied by x. This would be x squared plus, we can add these two coefficients because they're both on And now we can simplify this, and you might have beenĪble to go straight to this if you are familiar with Plus the a times the b, which is of course going to be ab. And then we're gonna have x times b, so we're multiplying each Not gonna skip any steps here just to see it this time. Out it would be equal to, you're going to have the x times the x which is going to be x squared. But if you were to multiply what we have on the right-hand side Multiplying binomials if any of this looks strange to you. It is let's just multiply these two binomials using a and b, and we've done this in previous videos. I'll highlight a andī in different colors. ![]() We know what a and b are? So let's work through this together now. Can we rewrite thisĮxpression as the product of two binomials where So I encourage you to pause the video and see if you can figure The product x plus a, that's one binomial, times x plus b, where we need to figure out Or to put it another way, I want to write it as And what I'd like to do in this video is I'd like to factor it as
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