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Current time:0:00Total duration:9:54

Welcome back. Where we left the last
problem, we essentially found ourselves having to solve
this equation right over here. And where we left off, we
said, well, maybe, just maybe, we might be able to
simplify this a little bit if we can find some common
factors between the numerator and the denominator. And I want to do it here because
if I multiplied this out, and multiply it times
the square root of 3, and then subtract
it or add it to that so I got just a straight-up
polynomial, that polynomial is going to have the
square root of 3 in it. It's going to be a really,
really messy thing. And it's still going
to be fourth degree. And at least this way,
I have some chance of being able to solve it. At least I just have integer
coefficients over here. And the only way that
this is going to simplify is if it has any common factors
with the stuff down here. And when you look
over here, this right here is actually going
to have complex roots if you did the negative b plus
or minus the square root of b squared minus 4ac over 2a. The b squared
minus 4ac over here is actually going
to be negative. So it's not going to
have any real roots. So that is not going to have any
easy, real roots, but this is. The x squared plus 1, that
has x plus 1 and x minus 1. Or another way to think
about it is roots to this are negative 1 and positive 1. So if either of those
are roots to this, then we're in business. And actually, just
looking at this, there's a decent chance,
because in general, if you want to think about the
roots of really any n-th degree polynomial, the roots,
if you think about them-- let me just write it over here. If I have a bunch of roots--
x minus r1, x minus r2, x minus r3, and
then let's say we have a fourth
degree, x minus r4. So let's say we have four roots. When you multiply
this whole thing, you're going to have x to the
fourth plus blah, blah, blah, blah, blah, all the way. And then the last term over here
is going to be r1, r2, r3, r4. It's going to be the
product of all of these. And so if you wanted
to put this equation right or this
expression in this form, you divide the whole
expression by 2. Or you can kind of say, well,
let's just look at this thing and divide it by 2. And then you would have
x to the fourth plus 2x, all of that plus 1/2. And so this thing would
have to, in this context, be equal to 1 over 2. And away, this is
kind of a-- it's not a fail-proof
technique, but it is one way to kind of
discover roots, is you, say, OK, the roots could
be the factors of this divided by the factors of that. And 1 is definitely a
factor of both of those. And actually, if you
think about it, negative 1 could be as well. So we feel good. But let's actually try it out. Instead of going through
all of this argument, let's actually see if
either 1 or negative 1 are factors of
this thing up here. If they're not,
then we just have to move ahead and multiply
this thing out, multiply it by square root of 3,
and then just solve. But we can hope. Usually the test writers
aren't that big sadists. Otherwise, it would be
impossible to solve. So I'm feeling good about this. So let's just try it with 1. So if we try 1 here,
so let's put 1 here and see if we get 0 over here. So if we get 2 times 1 to
the fourth plus-- so let me just write that. So 2 times 1 to the fourth
is just 2, plus 2 times 1 to the third is just 2, minus
3 times 1 squared is just 3, minus 2 plus 1. So this is 4 minus 5 plus 1. That indeed equals 0. And let's try negative 1. So this is 2 times
negative 1 to the fourth. That's still 2. Plus 2 times negative 1 to the
third, so that's negative 2. Minus 3 times negative 1
squared, that's negative 3. And then minus 2 times negative
1, so that's plus 2 plus 1. So this is 0 minus
3 plus 2 plus 1. This also cancels out. This also gets us 0. So we now know that both
x plus 1 and x minus 1 are factors of
this thing up here. Or another way to
think about it is their product is a factor
of this thing up here. x squared minus 1 is
a factor of this. So if we were to divide x
squared minus 1 into this, we should get a
reasonable answer. We shouldn't get a remainder,
I guess I should say. So let's do that. So we're going to get
a little practice doing polynomial long division. So if we divide x squared minus
1 into this thing up here, 2x to the fourth plus
2x to the third minus 3x squared minus 2x plus 1. So this is good practice on
polynomial long division-- highest degree term going
into highest degree term. x squared goes into 2x to
the fourth 2x squared times. So let's put that in
the 2x squared spot. 2x squared times x squared
is 2x to the fourth. 2x squared times negative
1 is negative 2x squared. And now let's subtract
this from up there. So we'll multiply by it
negative 1 and then add it. And so this gives
us 2x to the third. Negative 3 plus 2, so it's
minus x squared minus 2x plus 1. And then x squared goes into
2x to the third 2x times. So this is plus 2x. 2x times x squared
is 2x to the third. 2x times negative
1 is negative 2x. And then we can subtract
this from there, or you could say
add the negative. So let's just add the negative. It's easier to keep
track of things. So those cancel out. You get negative x squared--
these cancel out-- plus 1. x squared goes into negative
x squared minus-- or negative 1 times. Negative 1 times x squared
is negative x squared. Negative 1 times
negative 1 is positive 1. And indeed, we get no remainder. So we can write
this thing up here. We can write it as
the product of this. So it is equal
to-- so this thing we can rewrite as 2x squared
plus 2x minus 1 times x squared minus 1. So this times this is
equal to this, so times x squared minus 1. So we just factored it. And so that simplifies
things a good bit. It doesn't make
it trivial, but it simplifies things a good bit. So now we can cancel
that with that. And now at least we have
things in second degree. We got rid of these third
and fourth degree terms. So let's now rewrite things. So now let's multiply both
sides of this equation by x squared plus x plus 1. So we're going to get
the square root of 3-- so let me just write it out. So I'm going to get the
square root of 3 times x squared plus the
square root of 3 times x plus the square
root of 3 times 1-- so I just multiplied the
left-hand side by what we have left in the denominator-- is
equal to what we have left in the numerator,
because if we multiply the right side by this,
it obviously cancels out. So this is going to be equal
to 2x squared plus 2x minus 1. And now if we want to-- now
let's see what we can do here. Well, we can subtract the
stuff on the right side from the left-hand side. And so we would get the
square root of 3 minus 2 times x squared plus the square
root of 3 minus 2 times x-- just subtracting this
stuff from that side-- plus the square root
of 3 plus 1, right? Did I do that right? Square root of 3, I subtracted
the x squared from that, so square root of 3 minus
2, square root of 3 minus 2. And you have the
square root of 3. Subtracting a negative 1,
I get a plus 1 over here. And since I subtracted
all of this stuff from both sides
of the equation-- this is just the
left-hand side-- the right-hand side is now 0. I subtracted this from the
right-hand side to get 0. So now we have-- this almost
looks ridiculously simple. Even though I have these
ridiculous radicals and all that, I have a straight-up
quadratic equation right here, so I can use the
quadratic formula. So the roots here, so x is
going to be equal to negative b. This is b right here. So negative b, you
can swap those. So it's 2 minus the square root
of 3 plus or minus b squared. So let me just write it here. So the square root of 3 minus
2 squared minus 4 times a. So that's the square root
of 3 minus 2 times c. Well, c is the entire constant
term right over there. So it's times square root of
3 plus 1, all of that over 2a, so all of that over
this business over here. So that's 2 root 3 minus 4. So we thought we were
in the home stretch, but now we have to do some
major radical simplification. So once again,
these test writers are not pleasant people. So let's try our
best to tackle this. So let's go into-- let's
go inside the radical right over here. Let's go inside of the
radical right over here. And you have the square
root of 3 minus 2. Actually, I'm going
to leave you there. I'm approaching 10
minutes, and frankly, I need a drink of water. This problem is maybe the most
tiring problem I've ever done. I'll leave you there. We just have to simplify
this in the next video.