Dividing quadratics by linear expressions with remainders | Algebra 2 | Khan Academy

So if you've been watching these videos, you know that we have a lot of scenarios where people seem to be walking up to us on the street and asking us to do math problems, and I guess this will be no different.

So let's say someone walks up to you on the street and says, "Quick! ( \frac{x^2 + 5x + 8}{x + 2} ), what can this be simplified to?" Or, "What is ( \frac{x^2 + 5x + 8}{x + 2} )?"

Pause this video and see if you can work through that.

There are two ways that we can approach this. We can try to factor our numerator and see if we have a common factor there, or we can try to use algebraic long division.

So let's first try to factor this numerator, and we would ideally want ( x + 2 ) to be one of the factors. So let's see what two numbers can add up to 5, and when I multiply them, I get 8.

Ideally, 2 is one of them, so I can think of 2 and 3, but ( 2 \times 3 ) is going to be equal to 6, not 8. I can't think of anything else.

But that still gives us some progress, because what if we did say, "Alright, let's rewrite part of it"? What if we were to write ( x^2 + 5x ), and we want to write a + 6, because that actually would be divisible by ( x + 2 ).

So I'm going to write ( + 6 ), but of course, we have an 8 here, so then we're going to have an extra 2 right over there, and then all of that is divisible by ( x + 2 ).

Now I can rewrite this part up here in orange: that is ( x + 2 ) times ( x + 3 ). So let me write it here: ( x + 2 ) times ( x + 3 ). I still have that ( + 2 ) sitting out there in the numerator ( + 2 ), and then all of that over ( x + 2 ).

I could also write this as being over ( x + 2 ) and this being over ( x + 2 ). All I did is I said, "Hey, if I have something plus something else over ( x + 2 ), that could be the first something over ( x + 2 ) plus the second something over ( x + 2 )."

Here we can say, "Hey, look, this first part, as long as ( x ) does not equal negative 2, because then we would be changing the domain, these two would cancel out." You could say, "Hey, I'm just dividing the numerator and the denominator by ( x + 2 ), and so this would be equal to ( x + 3 + \frac{2}{x + 2} )."

I would have to constrain the domain, so this is for ( x ) does not equal negative 2.

In this situation, we had a remainder, and people will refer to the 2 as the remainder. We divide as far as we can, but we still have it left to divide the 2 by ( x + 2 ). So we would refer to the 2 as the remainder.

Now that wasn't too difficult, but it also wasn't too straightforward. We'll see that, and this is a situation where the algebraic long division is actually a little bit more straightforward.

So let's try that out, and once again pause this video and see if you can figure out what this is through algebraic long division.

We're trying to take ( x + 2 ) and divide it into ( x^2 + 5x + 8 ). Look at the highest degree terms: the ( x ) and the ( x^2 ). ( x ) goes into ( x^2 ) ( x ) times.

Put it in the first-degree column: ( x ) times 2 is ( 2x ) and ( x ) times ( x ) is ( x^2 ). Subtract these from ( x^2 + 5x ), and we get ( 5x - 2x = 3x ).

( x^2 - x^2 ) is just 0. Bring down that 8. Look at the highest degree term, and we get ( x ) goes into ( 3x ) 3 times. Put that in the constant column or the 0th-degree column, so ( + 3 ).

( 3 \times 2 = 6 ), ( 3 \times x = 3x ). Subtract these, and we are left with… let me scroll down a little bit. You're left with those canceling out, and you're left with ( 8 - 6 ), which is indeed equal to 2.

We could say, "Hey, we don't really know how to divide ( x + 2 ) into 2 for an arbitrary ( x ), so we will say, 'Hey, this is going to be equal to ( x + 3 ) with a remainder of 2.”

Once again, if you wanted to rewrite that original expression and you wanted it to be completely the same, including the domain, you would have to constrain the domain just like that.