Multiplying Binomials with Radicals

u16_l2_t3_we1 Multiplying Binomials with Radicals

We're asked to multiply
and to simplify.

And we have x squared minus
the principal square root

of 6 times x squared plus the
principal square root of 2.

And so we really just
have two binomials,

two two-term expressions
that we want to multiply,

and there's multiple
ways to do this.

I'll show you the
more intuitive way,

and then I'll show
you the way it's

taught in some
algebra classes, which

might be a little bit
faster, but requires

a little bit of memorization.

So I'll show you the
intuitive way first.

So if you have
anything-- so let's say

I have a times x
plus y-- we know

from the distributive
property that this

is the same thing as ax plus ay.

And so we can do the
same thing over here.

If you view a as x squared--
as this whole expression

over here-- x squared minus
the principal square root of 6,

and you view x plus y
as this thing over here,

you can distribute.

We can distribute all
of this onto-- let

me do it this way-- distribute
this entire term onto this term

and onto that term.

So let's do that.

So we get x squared minus
the principal square root

of 6 times this term-- I'll do
it in yellow-- times x squared.

And then we have plus
this thing again.

We're just distributing it.

It's just like they say.

It's sometimes
not that intuitive

because this is
a big expression,

but you can treat it just like
you would treat a variable over

here.

You're distributing it over
this expression over here.

And so then we have x squared
minus the principal square root

of 6 times the principal
square root of 2.

And now we can do the
distributive property again,

but what we'll do is we'll
distribute this x squared

onto each of these terms and
distribute the square root of 2

onto each of these terms.

It's the exact
same thing as here,

it's just you could imagine
writing it like this.

x plus y times a is still
going to be ax plus ay.

And just to see the pattern, how
this is really the same thing

as this up here,
we're just switching

the order of the multiplication.

You can kind of view it as we're
distributing from the right.

And so if you do this, you
get x squared times x squared,

which is x to the fourth,
that's that times that, and then

minus x squared times the
principal square root of 6.

And then over here you
have square root of 2 times

x squared, so plus x squared
times the square root of 2.

And then you have square root
of 2 times the square root of 6.

And we have a negative
sign out here.

Now if you take the
square root of 2--

let me do this on the
side-- square root of 2

times the square root of 6, we
know from simplifying radicals

that this is the exact same
thing as the square root of 2

times 6, or the principal
square root of 12.

So the square root of 2
times square root of 6,

we have a negative
sign out here,

it becomes minus the
square root of 12.

And let's see if we can
simplify this at all.

Let's see.

You have an x to
the fourth term.

And then here you
have-- well depending

on how you want to view
it, you could say, look,

we have to second degree terms.

We have something
times x squared,

and we have something
else times x squared.

So if you want,
you could simplify

these two terms over here.

So I have square
root of 2 x squareds

and then I'm going to subtract
from that square root of 6

x squareds.

So you could view this
as square root of 2

minus the square root of 6, or
the principal square root of 2

minus the principal square
root of 6, x squared.

And then, if you want,
square root of 12,

you might be able
to simplify that.

12 is the same
thing as 3 times 4.

So the square root of 12
is equal to square root

of 3 times square root of 4.

And the square root of 4, or
the principal square root of 4

I should say, is 2.

So the square root of
12 is the same thing

as 2 square roots of 3.

So instead of writing the
principal square root of 12,

we could write minus 2 times
the principal square root of 3.

And then out here you have
an x to the fourth plus this.

And you see, if you
distributed this out,

if you distribute this x
squared, you get this term,

negative x squared,
square root of 6,

and if you distribute it onto
this, you'd get that term.

So you could debate which
of these two is more simple.

Now I mentioned
that this way I just

did the distributive
property twice.

Nothing new, nothing fancy.

But in some classes, you will
see something called FOIL.

And I think we've done
this in previous videos.

FOIL.

I'm not a big fan of
it because it's really

a way to memorize a process
as opposed to understanding

that this is really just from
the common-sense distributive

property.

But all this is is
a way to make sure

that you're multiplying
everything times

everything when
you're multiplying

two binomials times
each other like this.

And FOIL just says, look,
first multiply the first term.

So x squared times x
squared is x to the fourth.

Then multiply the outside.

So then multiply--
I'll do this in green--

then multiply the outside.

So the outside terms are x
squared and square root of 2.

And so x squared times
square root of 2--

and they are positive-- so
plus square root of 2 times x

squared.

And then multiply the inside.

And you can see
why I don't like it

that much is because you
really don't know you're doing.

You're just applying
an algorithm.

Then you'll multiply the inside.

And so negative square
root of 6 times x squared.

And then you multiply
the last terms.

So negative square root of
6 times square root of 2,

that is-- and we
already know that-- that

is negative square root of
12, which you can also then

simplify to that expression
right over there.

So it's fine to use
this, although it's good,

even if you do use this, to
know where FOIL comes from.

It really just comes from
using the distributive property

twice.