University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Factoring a polynomial by factoring. So behind me I have a fourth degree polynomial. It’s a little bit disguised but we have the fourth degree in an x, we have 27x to the fourth equals 8x. So we’re trying to solve when is it equal to each other?
Now, the first thing I want to talk about is a common mistake that occurs and that people say, okay, I have an x on either of the sides, let’s divide by x. Now what actually happens is you loose a solution if you ever do that, okay? I want to do this a little quickly, we’ll do a little side now.
If I divide both sides by x, I end up with, I divide by x I end up with 27x³ is equal to 8. So these are fairly equivalent statements, those divided by x to get 1. But just looking at it, this first one, what’s the easiest solution to see that works? 0, if I have x is equal to 0, I plug 0 into 1, I get 0, 0 in the other, 0, zero is equal to 0. So 0 is the solution to this equation. By dividing by that variable, by dividing by x, what I have done is actually gotten rid of that answer as a possibility. If I plug in 0 now, I end up with 27 times 0, 0 is equal to 8, okay? So 0 was a solution at first, but by dividing by it I actually got rid of it.
So that’s a little lesson in never dividing by 0, okay? You can’t do it, you always have to bring things to the same side. So in this case I want to subtract 8x over instead of dividing by x, leaving me with 27x to the fourth, minus 8x is equal to 0. From here I want to try to factor this. In order to solve a polynomial like this I need to figure into multiplication instead of addition and subtraction.
So, looking at these, what do these have in common? X, factor that out, leave this with 27x³ minus 8 is equal to 0. So now we know that x can be zero or we have this statement here, 27x³ minus 8 is equal to 0 either.
Now, the instructions are solving this by factoring, okay? if we look at this we recognize that this is the difference of cubes, so we can factor that accordingly. This statement right here is actually quantity 3x³ and this is 2³.So factor this up using our difference of cube formulas. We have x, 3x minus 2, we then have 3x² so that turns into 9x², opposite sign, so that turns into a plus, 6x and then plus 4 and that is going to be equal to 0. Now we want to solve it out.
When we’re multiplying a bunch of things together to equal 0, each one could be 0, so that will tell us we have our x here, x could be 0, x could be 2/3 or lastly this could be equal to 0. However, something to note is that whenever we’re dealing with a difference of cubes, this statement that last trinomial, that last three terms won’t be factorable, okay? So we actually wont get any solutions out of this, it’s just a little extra term that goes along for the ride, it will never be 0 itself.
So what we end up with is just these two solution, 0 or 2/3. Always making sure to bring everything to one side and then factoring and never dividing by a variable.
Unit
Factoring