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Dividing Polynomials - Problem 2 3,329 views

Teacher/Instructor Alissa Fong
Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

I’m going to do this problem using long division, so the first thing I need to do is write it out long division style, meaning the divisor is going to go out and this whole big thing is called my dividend is going to go under this division sign. I’m going to show you the biggest mistake students make right now and the very, very first step when they’re rewriting this problem.

A lot of students just write this exactly the way it is. Do you see why that might be a problem? The reason why is because although this is in standard form, notice that I don’t have an x² term that’s totally going to mess up my long division algorithm that I learnt in fourth grade. So the way I want to write this instead of just rewriting it exactly, I’m going to stick in there 0x² as a place holder. It seems like a small deal right now, but it actually makes a huge difference when you’re going through this process.

Once I have that 0 in there, these problems are just pretty straight forward like the ones you’ve already done in your homework. Think about x times what gives me x to the third? It’s going to be x² and I’ll write it on top of my x² place holder, multiply and then make sure I subtract both of those terms so that my x to the thirds are eliminated and I’m going to have negative, or subtract minus 4x²which is the same thing as positive 4x². Bring down -13x. X² times 4x is going to give me that, multiply here I’m going through a little more quickly now because you guys I think are getting the hang of it.

Again the biggest mistake students make is they forget to distribute this negative sign. I have -13x's minus, minus which is the same thing as plus 16x's. Watch out. Okay and then I have plus 3. I think that’s the end. I’m going to go through and see if I have a remainder let’s see 3 times -4 is -12 yeah. Okay so I don’t have a remainder which is handy.

So when you’re going through to doing these problems, the most common mistake students are going to make in problems like these is the very first step. That 0x² place holder is absolutely critical when you’re setting up this Algorithm, so please when you’re rewriting you’re problem, double check and make sure not only is it in standard form, but look for a missing term and then stick that guy in there so you have a place holder.

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