Dividing polynomials by x (no remainders) | Algebra 2 | Khan Academy

What I'd like to do in this video is try to figure out what ( x ) to the fourth minus ( 2x ) to the third plus ( 5x ) divided by ( x ) is equal to. So pause this video and see if you can have a go at that before we work through this together.

All right, so if we're saying what is this top expression divided by this bottom expression, another way to think about it is what do I have to multiply? So I'm going to multiply something; I'll put that in parentheses. If I multiply that something times ( x ), I should get ( x ) to the fourth minus ( 2x ) to the third plus ( 5x ).

Now, how do I approach that? Well, there are two ways that I could tackle it. One way is I could just rewrite this expression as being, and I will just make this ( x ) in yellow so I can keep track of it. I could just rewrite this as ( 1 ) over ( x ) times ( x ) to the ( 4th ) minus ( 2x ) to the third plus ( 5x ), and then I can distribute the ( 1 ) over ( x ).

So what is that going to be equal to? What’s it going to be equal to ( x ) to the fourth over ( x ) minus ( 2x ) to the third over ( x ) plus ( 5x ) over ( x )? So what are each of these going to be equal to? ( x ) to the ( 4th ) divided by ( x ): if I have 4 ( x's ) that I'm multiplying together and then I divide by ( x ), that’s going to be equivalent to ( x ) to the third power. So this right over here is equal to ( x ) to the third.

You could also get there from your exponent properties; in the denominator, you have an ( x ) to the first power, and so you would subtract the exponents. You have the same base here, so that’s ( x ) to the third. And then this part right over here, what would that equal to? Well, it's going to be minus ( 2x ) to the third divided by ( x ) to the first. Well, by the same property, that's going to be ( x^2 ).

And then, last but not least, if you take five ( x's ) and then you divide by ( x ), you are just going to be left with five. You can verify that this indeed, if I were to multiply it by ( x ), I'm going to get ( x ) to the fourth minus ( 2x ) to the third plus ( 5x ).

Let me do that; if I put ( x ) to the third minus ( 2x^2 ) plus ( 5 ) times ( x ), what I can do is distribute the ( x ). ( x ) times ( x ) to the third is ( x^4 ), ( x ) times negative ( 2x^2 ) is negative ( 2x^3 ), ( x ) times ( 5 ) is ( 5x ).

Now, I mentioned there are two ways that I could do it. Another way that I could try to tackle it is I could look at this numerator and try to factor an ( x ) out. I would try to factor out whatever I see in the denominator. So if I do that, actually, let me just rewrite the numerator.

So I can rewrite ( x ) to the fourth as ( x ) times ( x^3 ), and then I can rewrite the minus ( 2x ) to the third as, let me write it this way, as plus ( x ) times negative ( 2x^2 ), and then I could write this ( 5x ) as being equal to plus ( x ) times ( 5 ).

Then I’m going to divide everything by ( x ). I just rewrote the numerator here, but for each of those terms, I factored out an ( x ). Now I can factor out ( x ) out of the whole thing. So I sometimes think of factoring out an ( x ) out of the whole thing as reverse distributive property.

So if I factor out this ( x ) out of every term, what am I left with? I'm left with ( x ) times ( x^3 - 2x^2 + 5 ). I ended up doing that in the wrong color, but hopefully, you're following—plus ( 5 ), and then all of that is divided by ( x ).

As long as ( x ) does not equal zero, ( x ) divided by ( x ) is going to be equal to one. And we're left with what we had to begin with, or the answer that we had to begin with.

So these are two different approaches. Nothing super sophisticated here. When you're dividing by ( x ), you're just like, "Hey, that's the same thing as multiplying every term by ( 1 ) over ( x )" or you can factor out an ( x ) out of the numerator, and then they cancel out.