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## Dividing polynomials by linear factors

Current time:0:00Total duration:4:32

# Dividing polynomials by linear expressions: missing term

CCSS Math: HSA.APR.D.6, HSA.APR.D

## Video transcript

In front of us, we have another screenshot from Khan Academy and I've
modified it a little bit so I have a little bit of extra space. And it says, divide the polynomials. The form of your answer should either be a straight up polynomial or a polynomial plus the remainder over x minus five, which we have here in the denominator, where p of x is a polynomial
and k is an integer. So, we've done stuff like this, but like always, I encourage
you to pause this video and work on this on your own. And if you were doing
this on Khan Academy, there's a little bit of an input box here where you'd have to type in the answer, but let's just do it on paper for now. All right. So, we're trying to figure out what x minus five divided into two x to the third power. Actually, I want to be careful here because I want to be very, very organized about my different degree columns. So, this is my third degree column. And then I want my second degree column. But there is no second degree term here. There's a first degree term. So, I'll write it out here. So, minus 47x. And actually, to be even more careful I'll write plus zero x squared. And then I have minus 15. By putting that plus zero x squared, that's making sure I'm doing good, I guess degree place column hygiene. All right, so now we
can work through this. And first we could say hey, how many times does x go into the
highest degree term here? Well, x goes into two x to the third power two x squared times. We'd want to put that in
the second degree column. Two x squared. You can see how it would've gotten messy if I'd put negative 47x here. I'd be like where do I
put that two x squared and you might confused yourself, which none of us would want to happen. All right, two x squared
times negative five is negative 10x squared. Two x squared times x is
2x to the third power. Now we want to subtract
what we have in red from what we have in blue. So, I'll multiply them
both by negative one. So that becomes a negative and then that one becomes a positive. And that's actually one
of the biggest areas for careless errors. If you had negative here and
you just want to subtract it because you know you have to subtract it. Be like no, I'm subtracting a negative, it needs to be a positive now. All right. So, zero x squared plus
10x squared is 10x squared. And then the two x to the third minus two x to the third is just zero. And then we can bring
down that negative 47x. And once again, we look at
the highest degree terms. X goes into 10x squared 10x times. So, plus 10x. 10x times negative five is negative 50x. 10x times x is 10x squared. And once again, we want to
subtract what we have in teal from what we have in red. So we could multiply both
of these times negative one. That becomes a negative. This one becomes a positive. Now, negative 47x plus
50x is positive three x. And then 10x squared minus
10x squared gets canceled out. Bring down that 15. Come on down. Used to watch a lot of
Price is Right growing up. Never quite made it to the show. All right. X goes into three x how many times? It goes three times. Three times negative five is negative 15. Three times x is three x. We want to subtract the
orange from the teal. And so, this becomes a negative,
this becomes a positive. Negative 15 plus 15 is zero. And three x minus three x is zero. So, you're just left with zero. So, no remainder. So this whole thing, you could
re-express or simplify as two x squared plus 10x plus three. And once again, if this
was on Khan Academy, there would be a little
bit of an input box that looks something like this and you would have to type this in. Now, if you wanted these to
be exactly the same expression you would also need to
constrain the domain. You would say, okay four x
does not equal positive five. And the reason why you
have to constrain that is the whole reason why we can
even divide by x minus five is we're assuming that x minus
five is no equal to zero. And it's generally not equal to zero as long as x does not
equal to positive five. But for the sake of that exercise, you don't need to constrain
the domain like this.