![]() That needs to be a timesī, right over there. And then a times b, right over here, that needs to be equal A plus b, needs to beĮqual to negative 20. See if we can factor this into an x plus a, and an x plus b. That is going to be equal to, that equals to x squared, plus a, plus b, x, plus a b. 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 So let's see if we can factor, if we can express this quadratic as a product of two expressions. Going to be left with x squared, and then negative 120, divided by six. If I do the same thing to both sides of the equation, then the equality still holds. If we divide the left sideīy six, divide by six, divide by six, divide by six. Of this equation by six, I'm still going to have It looks like actually all of these terms are divisible by six. The second degree term, on the x-squared term. To do is see if I can get a coefficient of one, on I might be able to deal with, and I might be able to factor, Alright, let's work through this together. Like always, pause this video, and see if you can solve for x, if you could find the x values X squared, minus 120 x, plus 600, equals zero. So don't feel bad that you couldn't factor by grouping-this isn't a good victim for that method. That means there is a factor of (h - 8), leaving h² + 12h + 84 as the other (quadratic) factor ![]() Now, with your h³ + 4h² - 12h - 672, if you graph this polynomial, there seem to be one positive root and two imaginary roots - the positive root is 8 So the factors would be (h + 4) and (h²- 12) and the roots would be -4, +2sqrt(3) and -2sqrt(3) So, possibly you wrote the example down wrong? Or if you made it up, you would have to have something like - 48 as the final term. (h + 4) is not actually a factor of this polynomial, but it would have to be in order for there to be a way for us to find it again in the second set of terms. So if you factor out the h², you get (h + 4) as you said.Īfter that, there is a problem with this method in this example. ![]() ![]() The tip-off is the 4 terms and the leading exponent of 3. If students can remember some simple generalizations about roots, they can decide where to go next.This looks like an example of factoring by grouping. Loh believes students can learn this method more intuitively, partly because there’s not a special, separate formula required. It’s quicker than the classic foiling method used in the quadratic formula-and there’s no guessing required. When solving for u, you’ll see that positive and negative 2 each work, and when you substitute those integers back into the equations 4–u and 4+u, you get two solutions, 2 and 6, which solve the original polynomial equation. When you multiply, the middle terms cancel out and you come up with the equation 16–u2 = 12. So the numbers can be represented as 4–u and 4+u. If the two numbers we’re looking for, added together, equal 8, then they must be equidistant from their average. Instead of starting by factoring the product, 12, Loh starts with the sum, 8. Those two numbers are the solution to the quadratic, but it takes students a lot of time to solve for them, as they’re often using a guess-and-check approach. “Normally, when we do a factoring problem, we are trying to find two numbers that multiply to 12 and add to 8,” Dr.
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