Professor Dave again, let’s practice simplifying expressions.
We just learned what roots and radical expressions are, as well as the basic rules for how to
manipulate them, so now it’s time to practice simplifying expressions containing radicals
and exponents. Let’s go through a few examples and highlight the strategies we must use in
each case. First let’s try something simple, six X cubed Y, quantity squared. When we say
the word quantity this way, it’s to distinguish between a whole term or set of terms undergoing
an operation rather than just the last thing mentioned. If I had simply said six X cubed
Y squared, this is what that language would imply, with the squared applying only to the
Y, but by inserting the word quantity, we can communicate the presence of these parentheses,
and the fact that all of this is being raised to this exponent. The first thing we must
understand about exponents that operate on parenthetical terms is that while they do
not distribute across sums or differences, they do distribute across products and quotients.
In other words, A plus B quantity squared is not equal to A squared plus B squared.
For example, five squared is twenty-five, but if we change five to three plus two and
try to distribute the exponent, three squared plus two squared is nine plus four, or thirteen,
so we clearly weren’t allowed to do what we did. To evaluate the square of a sum, we
need another technique that we will learn later. But A times B quantity squared is indeed
equal to A squared times B squared. This is easy to prove, because AB squared is AB times
AB, and since multiplication is associative, we can shuffle these around to get AABB, or
A squared B squared. So we can look at this term and rewrite it as six squared times X
cubed squared times Y squared. Now we continue to simplify. Six squared is six times six,
or thirty-six. For X cubed squared, let’s remember that X to the A to the B is equal
to X to the A times B, meaning that we multiply the two exponents together. That means X to
the three times two, which equals six, or X to the sixth. And Y squared stays as it
is. So we are left with thirty-six X to the sixth Y squared.
Let’s try a trickier one. We have four X cubed Y times this fraction in parentheses,
three X Y squared over two X cubed, all of which is squared. The first thing we want
to do is distribute this exponent over everything in the fraction. If the whole term is squared,
then each individual term can be squared. Three X Y squared becomes nine X squared Y
to the fourth, and two X cubed becomes four X to the sixth. Now we can combine like terms
for this term on the left and the numerator of this fraction. Remember that when multiplying
a fraction by a number, the number only multiplies the numerator, and this is no different. Four
times nine is thirty-six, X cubed times X squared is X to the fifth, and Y times Y to
the fourth is Y to the fifth. Now we can simplify a bit further, since thirty-six divided by
four is nine, and X to the fifth divided by X to the sixth is X to the negative one, since
five minus six is negative one. This is the same as leaving X in the denominator, because
X to the negative one is the same as one over X. So we are left with nine Y to the fifth
over X. How about this one, X to the sixth over 64
all raised to the negative two thirds power. There are a few ways to approach this, but
let’s first address the fact that the exponent is negative. If this were a single term, we
could put one over the term and then make the exponent positive, the way we said that
X to the negative one is one over X. But since we already have a fraction, let’s just flip
it over. If we turn this upside down, we can change the exponent to positive two thirds.
Now, as we recall, raising something to the two thirds power is the same as squaring it
and taking the cube root in some order. Since it’s easier to work with small numbers,
let’s take the cube root first, making sure to do it for both top and bottom. The cube
root of sixty-four is four. The cube root of X to the sixth, which is like X to the
sixth to the one third, is equal to X to the six times one third, which is two, so we get
X squared. That takes care of the denominator of the exponent, so now we just square the
top and bottom. That leaves us with sixteen over X to the fourth, which can’t be simplified
further. Let’s try another, ten A to the two X plus
Y power divided by two A to the X plus Y power. Here we can see that ten over two can be done
separately, which gives us five. Now to deal with this part of it, again we can just use
the rule for dividing some base raised to two different exponents; we just subtract
them. So A to the two X plus Y over A to the X plus Y gives us A to the two X plus Y minus
the quantity X plus Y. This only works because A is the base in both terms, if the bases
were different, we couldn’t do this. Then remember that we have to subtract the entire
exponent in the denominator, so we need parentheses to do this. The negative sign will then distribute,
leaving us with A to the two X plus Y minus X minus Y. Combining like terms, Y minus Y
goes away, and two X minus X leaves us with simply X. So the final answer is five A to
the X. How about this one. Three X squared root thirty-two Y squared minus four Y root
eighteen X to the fourth. First with the roots, we can’t take out thirty-two, but if we
make it sixteen times two, we can take out a four and a Y. We can combine like terms
to get twelve X squared Y root two. Again with the other term, eighteen doesn’t work,
but nine times two means we can take out a three and an X squared, and combining like
terms gives us twelve X squared Y root two. These terms are the same, and anything minus
itself is zero, so this whole expression actually simplifies to zero. I think by this point
we get the picture. Even when expressions seem very complicated, we just go one step
at a time, simplifying according to the rules we know for manipulating roots and exponents.
Now try a few on your own to check comprehension.