Solving quadratics by factoring: leading coefficient â   1 | High School Math | Khan Academy

So we have (6x^2 - 120x + 600 = 0). Like always, pause this video and see if you can solve for (x). If you can find the (X) values that satisfy this equation.

All right, let's work through this together. So 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^2) 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. So 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, 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^2), and then (-120 / 6) that is, let's see, (120) divided by (6) is (20), so that's (-20x).

Then (600) divided by (6) is (100), so plus (100) is equal to (0). Divided by (6) is equal to (0). So let's see if we can factor. If we can express this quadratic as the product of two expressions.

The way we think about this—and we've done it multiple times—is if we have something that is (x + a) times (x + b). This is hopefully a review for you; if you multiply that out, that is going to be equal to (x^2 + (a + b)x + ab).

So what we want to do is see if we can factor this into ( (x + a)(x + b) ). (A + B) needs to be equal to (-20) (that needs to be (a + b)), and then (a \times b) needs to be equal to the constant term (that needs to be (ab)).

So can we think of two numbers that, if we take their product, we get positive (100), and if we take their sum, we get (-20)? Well, since their product is positive, we know that they have the same sign. So they're both going to have the same sign; they're either both going to be positive, or they're both going to be negative.

Since we know that we have a positive product and since their sum is negative, well, they must both be negative. You can't add up two positive numbers and get a negative, so they both must be negative.

So let's think about it a little bit. What negative numbers, when I add them together, I get (-20), and when I multiply, I get (100)? Well, you could try to factor (100); you could say, well, (-2 \times -50) or (-4 \times -25), but the one that might jump out at you is (-10) times (-10).

And this is (-10 + -10), so in that case, both our (a) and our (b) would be (-10). We can rewrite the left side of this equation as ( (x - 10)(x - 10) ). Again, (x - 10) and that is going to be equal to zero.

All I've done is I've factored this quadratic, or another way, these are both the same thing as ( (x - 10)^2 = 0 ). So the only way that the left-hand side is going to be equal to zero is if (x - 10) is equal to zero.

You could think of this as taking the square root of both sides, and it doesn't matter if I take the positive or negative square root or both of them; it's the square root of (0).

So we would say that (x - 10) needs to be equal to zero, and so adding (10) to both sides, we have (x = 10) is the solution to this quadratic equation up here.