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

Let's say someone walks up to you on the street and they give you this expression:

x squared plus 7x plus 10 divided by x plus 2.

And they say, “See if you could simplify this thing.”

So, pause this video and see if you can do that.

One way to think about it is: what is x squared plus 7x plus 10 divided by x plus 2?

What is that going to be?

All right, now there are two ways that you could approach this. One way is to try to factor the numerator and see if it has a factor that is common to the denominator. So, let's try to do that.

We've done this many, many times. If this looks new to you, I encourage you to review factoring polynomials other places on Khan Academy.

But what two numbers add up to seven and when you multiply them you get ten?

Well, that would be two and five.

So we could rewrite that numerator as (x + 2)(x + 5).

And then, of course, the denominator you still have (x + 2).

Then we clearly see we have a common factor.

As long as x does not equal negative 2, because if x equals negative 2, this whole expression is undefined.

This is because then you get a 0 in the denominator.

So as long as x does not equal negative 2, then we can divide the numerator and the denominator by (x + 2).

Once again, the reason why I put that constraint is we can't divide the numerator and denominator by zero.

So for any other values of x, this (x + 2) will be non-zero, and we could divide the numerator and the denominator by that and they would cancel out.

We would just be left with x + 5.

So another way to think about it is that our original expression could be viewed as x + 5 for any x that is not equal to negative 2.

Now, the other way that we could approach this is through algebraic long division, which is very analogous to the type of long division that you might remember from, I believe, it was fourth grade.

So what you do is you say, “All right, I'm going to divide (x + 2) into (x squared + 7x + 10)."

In this technique, you look at the highest degree terms.

So you have an x there and an x squared there, and you say, “How many times does x go into x squared?”

Well, it goes x times.

Now you would write that in this column because x is just x to the first power.

You could view this as the first degree column.

It's analogous to the place values that we talk about when we first learn numbers or how we regroup or talk about place value.

But here you can view it as degree places or something like that.

Then you take that x and you multiply it times this entire expression.

So x times 2 is 2x.

Put that in the first degree column.

x times x is x squared.

What we want to do is subtract these things in yellow from what we originally had in blue.

So we could do it this way, and then we will be left with:

7x minus 2x is 5x, and then x squared minus x squared is just zero.

Then we can bring down this plus 10.

Once again we look at the highest degree term.

x goes into 5x five times.

That's a zero degree; it's a constant.

So I'll write it in the constant column.

5 times 2 is 10, and 5 times x is 5.

Then I'll subtract these from what we have up here, and notice we have no remainder.

What's interesting about algebraic long division, we'll probably see in another video or two, is that you can actually have a remainder.

Those are going to be situations where just the factoring technique alone would not have worked.

In this situation, this model would have been easier.

But this is another way to think about it:

You say, “Hey, look, (x + 2)(x + 5) is going to be equal to this.”

Now, if you wanted to rewrite this expression the way we did here and say, “Hey, this expression is equal to x + 5,” we would have to constrain the domain.

You'd say, “Hey, for all x's not equaling negative 2 for these to be completely identical expressions.”