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In this video, we're going to learn to divide polynomials,

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and sometimes this is called algebraic long division.

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But you'll see what I'm talking about

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when we do a few examples.

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Let's say I just want to divide 2x plus 4 and

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divide it by 2.

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We're not really changing the value.

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We're just changing how we're going to express the value.

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So we already know how to simplify this.

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We've done this in the past. We could divide the numerator

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and the denominator by 2, and then this

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would be equal to what?

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This would be equal to x plus 2-- let me write it this way--

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it would be equal to, if you divide this by 2,

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it becomes an x.

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You divide the 4 by 2, it becomes a 2.

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If you divide the 2 by 2, you get a 1.

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So this is equal to x plus 2, which is pretty

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straightforward, I think.

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The other way is you could have factored a 2 out of here,

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and then those would have canceled out.

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But I'll also show you how to do it using algebraic long

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division, which is a bit of overkill for this problem.

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But I just want to show you that it's not fundamentally

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anything new.

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It's just a different way of doing things, but it's useful

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for more complicated problems.

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So you could have also written this as 2 goes into 2x plus 4

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how many times?

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And you would perform this the same way you would do

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traditional long division.

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You'd say 2-- you always start with the highest degree term.

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2 goes into the highest degree term.

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You would ignore the 4.

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2 goes into 2x how many times?

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Well, it goes into 2x x times and you put

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the x in the x place.

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x times 2 is 2x.

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And just like traditional long division, you now subtract.

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So 2x plus 4 minus 2x is what?

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It's 4, right?

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And then 2 goes into 4 how many times?

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It goes into it two times, a positive two times.

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Put that in the constants place.

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2 times 2 is 4.

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You subtract, remainder 0.

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So this might seem overkill for what was probably a

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problem that you already knew how to do and

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do it in a few steps.

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We're now going to see that this is a very

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generalizable process.

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You can do this really for any degree polynomial dividing

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into any other degree polynomial.

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Let me show you what I'm talking about.

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So let's say we wanted to divide x plus 1 into x squared

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plus 3x plus 6.

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So what do we do here?

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So you look at the highest degree term here, which is an

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x, and you look at the highest degree term here,

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which is an x squared.

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So you can ignore everything else.

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And that really simplifies the process.

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You say x goes into x squared how many times?

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Well, x squared divided by x is just x, right? x goes into

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x squared x times.

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You put it in the x place.

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This is the x place right here or the x to

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the first power place.

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So x times x plus 1 is what?

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x times x is x squared.

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x times 1 is x, so it's x squared plus x.

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And just like we did over here, we now subtract.

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And what do we get?

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x squared plus 3x plus 6 minus x squared-- let me be very

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careful-- this is minus x squared plus x.

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I want to make sure that negative sign only-- it

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applies to this whole thing.

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So x squared minus x squared, those cancel out.

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3x, this is going to be a minus x.

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Let me put that sign there.

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So this is minus x squared minus x, just to be clear.

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We're subtracting the whole thing.

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3x minus x is 2x.

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And then you bring down the 6, or 6 minus 0 is nothing.

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So 2x plus 6.

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Now, you look at the highest degree term, an x and a 2x.

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How many times does x go into 2x?

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It goes into it two times.

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2 times x is 2x.

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2 times 1 is 2.

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So we get 2 times x plus 1 is 2x plus 2.

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But we're going to want to subtract this from this up

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here, so we're going to subtract it.

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Instead of writing 2x plus 2, we could just write negative

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2x minus 2 and then add them.

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These guys cancel out.

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6 minus 2 is 4.

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And how many times does x go into 4?

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We could just say that's zero times, or we could say that 4

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is the remainder.

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So if we wanted to rewrite x squared plus 3x plus 6 over x

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plus 1-- notice, this is the same thing as x squared plus

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3x plus 6 divided by x plus 1, this thing divided by this, we

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can now say that this is equal to x plus 2.

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it is equal to x plus 2 plus the remainder divided by x

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plus 1 plus 4 over x plus 1.

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This right here and this right here are equivalent.

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And if you wanted to check that, if you wanted to go from

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this back to that, what you could do is multiply this by x

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plus 1 over x plus 1 and it add the two.

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So this is the same thing as x plus 2.

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And I'm just going to multiply that times x plus

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1 over x plus 1.

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That's just multiplying it by 1.

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And then to that, add 4 over x plus 1.

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I did that so I have the same common denominator.

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And when you perform this addition right here, when you

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multiply these two binomials and then add the 4 up here,

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you should you get x squared plus 3x plus 6.

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Let's do another one of these.

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They're kind of fun.

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So let's say that we have-- we want to simplify x squared

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plus 5x plus 4 over x plus 4.

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So once again, we can do our algebraic long division.

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We can divide x plus 4 into x squared plus 5x plus 4.

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And once again, same exact process.

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Look at the highest degree terms in both of them. x goes

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into x squared how many times?

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It goes into it x times.

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Put it in the x place.

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This is our x place right here.

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X times x is x squared.

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x times 4 is 4x.

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And then, of course, we're going to want to subtract

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these from there.

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So let me just put a negative sign there.

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And then these cancel out.

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5x minus 4x is x.

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4 minus 0 is plus 4.

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x plus 4, and then you could even see this coming.

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You could say x plus 4 goes into x plus 4 obviously one

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time, or if you were not looking at the constant terms,

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you would completely just say, well, x goes

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into x how many times?

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Well, one time.

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Plus 1.

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1 times x is x.

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1 times 4 is 4.

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We're going to subtract them from up here, so it cancels

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out, so we have no remainder.

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So this right here simplifies to-- this is

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equal to x plus 1.

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And there's other ways you could have done this.

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We could have tried to factor this numerator.

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x squared plus 5x plus 4 over x plus 4.

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This is the same thing as what?

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We could have factored this numerator as x plus 4

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times x plus 1.

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4 times 1 is 4.

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4 plus 1 is 5, all of that over x plus 4.

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That cancels out and you're left just with x plus 1.

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Either way would have worked, but the algebraic long

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division will always work, even if you can't cancel out

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factors like that, even if you did have a remainder.

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In this situation, you didn't.

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So this was equal to x plus 1.

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Let's do another one of these just to make sure that you

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really-- because this is actually a very, very useful

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skill to have in your toolkit.

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So let's say we have x squared-- let me

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just change it up.

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Let's say we had 2x squared-- I could really make these

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numbers up on the fly.

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2x squared minus 20x plus 12 divided by-- actually, let's

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make it really interesting, just to show you that it'll

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always work.

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I want to go above quadratic.

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So let's say we have 3x to the third minus 2x squared plus 7x

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minus 4, and we want to divide that by x squared plus 1.

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I just made this up.

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But we can just do the algebraic long division to

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figure out what this is going to be or what this

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simplified will be.

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x squared plus 1 divided into this thing up here, 3x to the

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third minus 2x squared plus 7x minus 4.

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Once again, look at the highest degree term.

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x squared goes into 3x to the third how many times?

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Well, it's going to go into it 3x times.

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You multiply 3x times this, you get 3x to the third.

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So it's going to go into it 3x times.

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So you have to write the 3x over here in the x terms. So

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it's going to go into it 3x times, just like that.

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Now let's multiply.

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3x times x squared is 3x to the third, right?

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3x times squared plus 3x times 1.

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So we have a 3x over here.

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I'm making sure to put it in the x place.

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And we're going to want to subtract them.

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And what do we have?

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What do we have when we do that?

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These cancel out.

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We have a minus 2x squared.

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And then 7x minus-- because I just subtracted 0 from there--

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7x minus 3x is plus 4x, and we have a minus 4.

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Once again, look at the highest degree term.

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x squared and a negative 2x squared.

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So x squared goes into negative 2x

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squared negative 2 times.

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Negative 2, put it in the constants place.

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Negative 2 times x squared is negative 2x squared.

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Negative 2 times 1 is negative 2.

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Now, we're going to want to subtract these from there, so

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let's multiply them by negative 1, or

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those become a positive.

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These two guys cancel out.

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4x minus 0 is-- let me switch colors-- 4x minus 0 is 4x.

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Negative 4 minus negative 2 or negative 4 plus 2 is equal to

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negative 2.

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And then x squared, now it has a higher degree than 4x, and

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the highest degree here, so we view this as the remainder.

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So this expression we could rewrite it as being equal to

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3x minus 2-- that's the 3x minus 2-- plus our remainder

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4x minus 2, all of that over x squared plus 1.

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Hopefully, you found that as fun as I did.