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# Solving quadratics by factoring: leading coefficient ≠ 1

## Video transcript

- [Voiceover] We have six x squared, minus 120 x, plus 600, equals zero. Like always, pause this video, and see if you can solve for x, if you could find the x values that satisfy this equation. Alright, let's work through this together. The numbers here don't seem like outlandish numbers. They seem like something that I might be able to deal with, and I might be able to factor, so let's try to do that. The first thing I like to do is see if I can get a coefficient of one, on the second degree term, on the x-squared term. It looks like actually all of these terms are divisible by six. So if we divide both sides of this equation by six, I'm still going to have nice integer coefficients. Let's do that. Let's divide both sides by six. If we divide the left side by six, divide by six, divide by six, divide by six. And I divide the right side by six. If I do that, and clearly if I do the same thing to both sides of the equation, then the equality still holds. On the left-hand side, I am going to be left with x squared, and then negative 120, divided by six. That is, let's see. 120 divided by six is 20. So that's minus 20 x. Then 600 divided by six, is 100. So plus 100, is equal to zero divided by six. Is equal to zero. So let's see if we can factor, if we can express this quadratic as a product of two expressions. The way we think about this, and we've done it multiple times, if we have something, if we have x plus a, times x plus b, and this is hopefully review for you. If you multiply that out, that is going to be equal to, that equals to x squared, plus a, plus b, x, plus a b. What we want to do is see if we can factor this into an x plus a, and an x plus b. A plus b, needs to be equal to negative 20. That needs to be a plus b. And then a times b, right over here, that needs to be equal to the constant term. That needs to be a times b, right over there. Can we think of two numbers, that if we take their product, we get positive 100, and if we take their sum, we get negative 20? Well since their product is positive, we know that they have the same sign. They're both going to have the same sign. They're either both going to be positive, or their both going to be negative, since we know that we have a positive product. Since their sum is negative, well they both must both be negative. You can't add up two positive numbers, and get a negative. So they both must be negative. Let's think about it a little bit. What negative numbers, when I add them together I get negative 20, when I multiply it, I get 100? Well you could try to factor 100. You could say, well negative two times negative 50, or negative four times negative 25. But the one that might jump out at you is this is negative 10, times, I'll write it this way, negative 10, times negative 10, and this is negative 10, plus negative 10. So in that case, both our a and our b, would be negative 10. And so we can rewrite the left side of this equation as, we can rewrite it as, x, and I'll write it this way at first, x plus negative 10, times, x plus negative 10 again. X plus negative 10, and that is going to be equal to zero. So all I've done is I've factored this quadratic. Another way, these are both the same thing as x minus 10. I could rewrite this as x minus 10, squared, is equal to zero. The only way that the left-hand side is going to be equal to zero, is if x minus 10 is equal to zero. You could think of this as taking the square root of both sides. It doesn't matter if you're taking the positive or negative square root, or both of them. The square root of zero, is zero. So, we would say, that x minus 10 needs to be equal to zero. So x , adding 10 to both sides of this, you have x is equal to 10, is the solution to this quadratic equation, up here.