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Current time:0:00Total duration:6:02

CCSS Math: HSA.APR.D.7, HSA.APR.D

- [Voiceover] We're told
the following equation is true for all real values of y for which the expression
on the left is defined and D is a polynomial expression. And they have this equation here. What is D? All right. So essentially,
what they're saying is they don't want us to
somehow solve this equation. They're saying D is going to be some type of a polynomial expression. They tell us that right over there: D is a polynomial expression. And if we figure out what D is, this left-hand side of the expression is going to evaluate to one
for all real values of y for which the expression is defined. So let's think about how
we would tackle that. Well the first thing
that pops into my mind, if I'm dividing by a fraction
or a rational expression, that's the same thing as
multiplying by the reciprocal. So let's just rewrite this
on the left-hand side. So this is 20 y squared minus 80 over D times four, oh, let me do the reciprocal,
let me be careful, times the reciprocal of this. If I divide by something, the same thing as multiplying
by the reciprocal, so let me just swap the
numerator and the denominator, (chuckle) numerators and denominators, numerator and denominator. All right, y to the third
plus nine y squared, all of that over four y
squared minus eight y. That's going to be equal to one. Now let's say if we can simplify all of these business on the
left-hand side a little bit. So let's see. Over here, I can divide both terms by 20. So let me factor out 20 because I think it's gonna end up being
a difference of squares. So if I factor out of 20, so this is about the same thing as 20 times y squared minus four and y squared minus four, we can rewrite as y plus
two times y minus two. It is a difference of squares. So let me write y plus two times y minus two. All right. This down here,
four y squared minus eight y, well it looks like we
can factor out of four y. And so this is going to be same thing as four y times y minus two. All right. So let me cross that out. So it's the same thing as
four y times y and minus two and I already see that
this y minus two here and this y minus two here
are gonna cancel out. And let's see. Up here, both terms are divisible by y squared,
so I can rewrite this as, I don't know if this is
actually going to be helpful because it's gonna, well let me just do it, just in case. So that's the same thing
as y squared times, y squared times y plus nine. All right. And so we can
rewrite all of these things. If we were to multiply
everything together, we would end up getting, the numerator would
get 20 times y plus two times y minus two times y squared times y plus nine, I'm just
multiplying all the numerators, and that's going to be over, in the denominator, I would have whatever the expression D is, times four y times y minus two, and that's all going to be equal to one. Now let's think about it. We can divide, we have y minus two
divided by y minus two, so those cancel out. Now let's see. We have a... We can divide the numerator
and the denominator by y, so that would just become one. And then that would just
become a y to the first power. And so, what we'd be left with, what we'd be left with in the numerator is 20 times this y times y plus two times y plus nine over, over four D is equal to one. Now if we wanna solve four D, well we could just
multiply both sides by D and one times D is just gonna be D, so we're gonna have D
equals something over here and we'd be done. So let's do that. D times that and let's multiply that times D. Let me be clear what I'm doing here. Let me draw a little divider here to make it clear that that's happening in the other side of the page. All right. So D times
this, those cancel out, and we're gonna be left with 20 y times, actually, let me, I could
simplify it even more. 20 divided by four is five. So, our numerator is now just one. So we have five times y times y plus two times y plus nine is equal to D, and we're done. This is D. This is the polynomial, whoops, that is... That is the polynomial expression that we are looking for. If you were to substitute this back in and then try to simplify it, well, you would end up
with all of these over here and D would be this, and so, it would all just cancel out and you would be left with one. For all real values y
for which the expression on the left is actually defined. There are some values of y for which the expression
on the left is not defined. If y is equal to zero,
this denominator is zero and you're dividing by zero,
well that's not defined. And then when you multiply
by the reciprocal, if this were to become zero, then that wouldn't be cool either. And there's multiple ways
to make this equal to zero. y could be equal to negative nine, that would also make this bottom zero. So we could think about
that if we wanted to, but they're not asking us to do that. They're saying, for all
of the real values for y for which the expression defined, find the D that makes all of
this business equal to one and we just did that.