Factoring using polynomial division: missing term | Algebra 2 | Khan Academy

We're told the polynomial ( p(x) ) which is equal to this has a known factor of ( x + 6 ). Rewrite ( p(x) ) as a product of linear factors. Pause this video and see if you can have a go at that.

All right, now let's work on this together. Because they give us one of the factors, what we can do is say, "Hey, what happens if I divide ( x + 6 ) into ( p(x) )? What do I have left over?" It looks like I'm still going to have a quadratic, and then I'll probably have to factor that somehow to get a product of linear factors. So let's get going.

If I were to try to figure out what ( x + 6 ) divided into ( x^3 + 9x^2 ), and now we're going to have to be careful. You might be tempted to just write -108 there, but then this gets tricky because you have your third-degree column, your second-degree column, you need your first-degree column, but you just put your zero-degree, your constant column here.

So to make sure we have good hygiene, we could write ( + 0x ), and I encourage you to actually always do this if you're writing out a polynomial so that you don't skip that place, so to speak, -108.

And so then you say, "All right, let's look at the highest degree terms." ( x ) goes into ( x^3 ) ( x^2 ) times. ( x^2 ) times ( 6 ) is ( 6x^2 ). ( x^2 ) times ( x ) is ( x^3 ). We want to subtract. We've done this multiple times, so I'm going a little bit faster than normal. Those cancel out.

( 9x^2 - 6x^2 = 3x^2 ). Bring down that ( 0x ). And then how many times does ( x ) go into ( 3x^2 )? Well, it goes ( 3x ) times, and we would write it in this column. Notice if we didn't keep this column for our first-degree terms, we'd be kind of confused where to write that ( 3x ) right about now.

And so ( 3x ) times ( 6 ), I should say, is ( 18x ). ( 3x ) times ( x ) is ( 3x^2 ). We want to subtract what we have in that, I guess that color is move light purple, not sure. And so we get ( 3x^2 )'s cancel out, and then ( 0x - 18x = -18x ). Bring down that ( -108 ).

And so then we have ( x ) goes into ( -18x ) ( -18 ) times. ( -18 ) times ( 6 ) is ( -108 ). That's working out nicely. ( -18 ) times ( x ) is ( -18x ), and then we want to subtract what we have in this not so pleasant brown color.

And so I will multiply them both by negative, and so I am left with zero; everything just cancels out. And so I can rewrite ( p(x) ). I can rewrite ( p(x) ) as being equal to ( x + 6 \times (x^2 + 3x - 18) ).

But I'm not done yet because this is not a linear factor; this is still quadratic. So let's see, can I think of two numbers that add up to ( 3 ) and then when I multiply I get ( -18 )? So they'll need different signs, and then the obvious one is positive ( 6 ) and negative ( 3 ).

And if that what I just did seems like voodoo to you, I encourage you to review factoring polynomials. But this I can rewrite because negative ( 6 + ) or actually I should say positive ( 6 + (-3) ) is equal to ( 3 ), and then positive ( 6 \times negative ( 3 ) is equal to ( -18 ).

So I can rewrite this as ( x + 6 \times (x + 6) \times (x - 3) ). And so there we have it; we have a product of linear factors, and we are done.