Brightstorm is like having a personal tutor for every subject
See what all the buzz is aboutCheck it out
Dividing Polynomials - Problem 3 2,901 views
We Math teachers spend so much energy trying to convince you that Math actually is sometimes used in real life. One time it can be used is when you’re looking at dimensions of some kind of cube or rectangular cube. You’ll learn more a lot about this in Geometry, but for now let’s look at this from an Algebra standpoint.
The volume of a room can be represented by v(x) equals x to the third plus x² take away 6x where the length is x take away 2. Find the expression for the width and height? Okay so this can be really intimidating for an Algebra student especially because you’re not officially in a Geometry class, but you’re having to do like a Geometry problem, but don’t worry this is just based on your Algebra skills.
First thing I always do is draw a picture because I’m a visual person and it helps me if I can like look at something and see how it relates to the problem. So we have a room that’s not the best picture, but you get the idea and we know that the volume is represented by that whole big polynomial, so I’m going to same volume equals x to the third plus x² take away 6x. I also know from my previous Geometry learning or my previous Math classes that the way to find the volume of a rectangle is to do length times width times height.
So this thing is equal to length times width times height. This whole big trinomial is equal to length times width times height and they also already told me the length is x take away 2, so really my volume is equal to x take away 2 times width times height. My job is to find those. If you’re looking at the picture, you can label length and width and height really however you want to. Height is usually the up and down part. Length I’m going to call x minus 2 and width will be there. I need to find those 2 values.
Okay well the way I’m going to find out what my width and height are are by using long division. So I’m going to back up a little bit and set up my long division problem where I’ll have x take away 2 going into x to the third plus x² minus 6x and then I’m going to use a place holder of 0 right there for my constant term, because I might have some kind of remainder otherwise well let’s see.
Okay x times x² is going to give me x³, then what I’m going to do is subtract both of these terms so that I have x² minus -2x² which is like plus. X² plus 2x² is 3x² minus 6x, now I need to think x times what gives me 3x² well I need a 3 and an x. Multiply again 3x times -2 is -6 when I subtract I find I have a remainder of 0.
What this tells me if I were to write this out is that x minus 2 times x² plus 3x equals that. I’m going to rewrite it over here. X minus 2 that was my length times x² plus 3x that was my quotient that I just found is equal to that volume expression. X to the third plus x² take away 6x. That’s a whole bunch of moving back and forth between letter and words, so let me just double check to make sure you’re clear on what I just did.
This was my volume, that’s the length I just found and this quotient that I found represents width times height. So I know I need to do something with this so that I can break it into 2 factors width and height. Well luckily this is easily factored. X multiplies into both of those terms so I’m going to rewrite this as x times x plus 3.
Now I found my width and my height. Actually I’m not sure which one is width and which one is height I don’t have to specify because the problem didn’t ask me to, but I know that one of my width or height is equal to x, the other one is equal to x plus 3 and the way I found that was by doing long division with my volume divided by my length expression to get a product that represented width times height and then that’s what I labeled here as my final answer.
The way I could check would be to go through and multiply this out and then take my result, multiply it by x minus 2 and that should give me my original volume statement. So guys don’t freak out if you see a Geometry problem and you’re only in an Algebra class. You can still do them most of the time they’re based on skills you’ve already learned in your Math classes.