So, we would say, that x minusġ0 needs to be equal to zero. You're taking the positive or negative square root, or both of them. You could think of this as taking the square root of both sides. The only way that the left-hand side is going to be equal to zero, is if x minus 10 is equal to zero. I could rewrite thisĪs x minus 10, squared, is equal to zero. So all I've done is I'veįactored this quadratic. X plus negative 10, and that is going to be equal to zero. 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. Times, I'll write it this way, negative 10, times negative 10, and this is negative 10, plus negative 10. But the one that might jump out at you is this is negative 10, Two times negative 50, or negative four times negative 25. I multiply it, I get 100? Well you could try to factor 100. When I add them together I get negative 20, when Since their sum is negative, well they both must both be negative. They're either both going to be positive, or their both going to be negative, since we know that we They're both going to have the same sign. If we take their sum, we get negative 20? Well since their product is positive, we know that they have the same sign. That if we take their product, we get positive 100, and 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. This looks like an example of factoring by grouping.
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