We're asked to multiply
the given radical expressions.
Notice in each product, we have a single radical expression
times a similar or difference of terms,
which means to determine the product,
we will need to distribute.
So for number one, to determine the product,
we distribute the square root of x,
which gives us two products.
This is equal to the square root of x times
two square root of x minus the square root of x
times three.
Now we determine the products using the rules
for multiplying radicals shown here.
To determine the product of the square root of x
and two square root of x,
we multiply xs under the square roots
or we multiply the radicands.
This equals two times the square root of x times x.
And then we have minus the square root of x times three
equals three square root of x.
Now we simplify the square roots.
Notice how this square root of x times x will simplify
because the radicand is a perfect square.
Let's try this one more time using exponents
to emphasize x times x is a perfect square.
Let's write this as two times the square root of x squared
minus three square root of x.
Now we simplify.
The square root of x squared simplifies perfectly
to one factor of x.
This equals two x minus three square root of x.
But because the index is on the square root is even,
the index is two and the exponent on the simplified variable
is odd, because x equals x to the first,
we must include an absolute value around the x.
The product is equal to two times the absolute value of x
minus three square root of x.
Next to determine the product of number two,
we distribute four square root y.
Once again, we have two products.
We have four square root y times five square root xy cubed
minus four square root y times the square root of y cubed.
To determine this first product, we multiply the numbers
outside the square roots then we multiply the radicands.
Four times five is equal to 20.
So we have 20 square root of y times xy cubed
equals xy to the fourth.
Then we have minus four square root y times
the square root of y cubed is equal to four times
the square root of y to the fourth.
And now we simplify both square roots
by identifying the perfect square factors of the radicand.
Notice y to the fourth is a perfect square,
because y to the fourth is equal
to y squared times y squared.
So let's write this as 20 times the square root of x
times y squared times y squared minus four
times the square root of y squared times y squared.
Again, y squared times y squared is a perfect square here
as well as here.
To emphasize this further, let's use the power property
of exponents and write y squared times y squared
as y to the second squared.
This is equal to 20 times the square root of x
times y to the second squared minus four times
the square root of y to the second squared.
And now we simplify.
The square root of y to the second squared simplifies
perfectly to one factor of y squared.
This is equal to 20 y squared times the square root of x
minus, again the square root of y to the second squared
simplifies to one factor of y to the second,
so we have minus four y to the second
or minus four y squared.
In this expression no absolute value's required,
because, while the index is even,
the exponent on the simplified variables
are even and not odd.
And now to determine the last product.
In this product, when we distribute, we'll have total
of one, two, three products.
We have the cubed root of z
times the cube root of z squared
minus the cube root of z times seven times
the cube root of z to the fifth,
and finally we have plus the cube root of z
times two times the cube root of z to the eighth.
For this first product we multiply the radicands
which gives us the cube root of z cubed minus,
here we have seven, then we multiply the radicands again,
which gives us the cube root of z to the sixth,
then we have plus two,
then we multiply the radicands which gives us
the cube root of z to the ninth.
And now we simplify the cube roots
by identifying the perfect cube factors of the radicands.
Well first we have the cube root of z cubed.
Z cubed is obviously a perfect cube.
So this will simplify perfectly.
Next we have minus seven
times the cube root of z to the sixth.
But z to the sixth is also a perfect cube,
because it's equal to z squared
times z squared times z squared.
This will also simplify perfectly.
And finally we have plus two times
the cube root of z to the ninth.
But z to the ninth is also a perfect cube,
because it's equal to z to the third
times z to the third times z to the third.
So, once again, this will also simplify perfectly.
To emphasize this even more,
let's rewrite this as the cube root of two to the third
minus seven times the cube root of z squared
raised to the third power.
Notice how this is still z to the sixth.
Then we have plus two times the cube root,
let's write this product as z to the third, cubed,
or raised to the third power.
And now we simplify.
The cube root of z to the third simplifies perfectly
to one factor of z, so we have z minus seven
times the cube root of z squared raised to the third
simplifies perfectly to one factor of z squared,
then we have plus two times
the cube root of z to the third cubed simplifies perfectly
to one factor of z cubed.
This is our final product.
And because the index is odd, no absolute value is required.
I hope you found this helpful.