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Homework Help: Math: Algebra: Exponents of Parenthesis
The previous lesson explained how to
simplify exponents of a single term inside parentheses, like the problem below.
(x3y4)5
This lesson covers how to simplify exponents on parentheses that contain a
polynomial (more than one term), like the problem below.
(x3 + y4)2
Because the two terms inside parentheses are not being multiplied or divided,
the exponent outside the parentheses can not just be "distributed in". Instead,
a 1 must be multiplied by the entire polynomial the number of times indicated
by the exponent. In this problem the exponent is 2, so it is multiplied two
times:
1(x3 + y4)(x3 + y4)
Use the FOIL Method
to simplify the multiplication above, then combine like terms.
x6 + x3y4 + x3y4 + y8
x6 + 2x3y4 + y8
Multiplication and Division
Examine the problem below.
(x3y4)5
Recall that multiplication is implied when there is no sign between a variable or set of parentheses and a number,
another variable, or another set of parentheses. Therefore in this problem,
the x3 and y4 are being multiplied.
In the next problem the x2 and x are being multiplied. The difference
is that a * is present which explicitly indicates multiplication. We will
solve this problem, then return to the first problem on the page.
(x2 * x)3
Because there is no addition or subtraction inside the parentheses,
the exponent can be just "distributed" in and simplified:
(x2*3 * x3)
x6 * x3
x9
Notice that this gives the same result as if we had simplified the inside of
the parentheses first, as we have done below.
(x2 * x)3
(x3)3
x3*3
x9
So why are there two different methods of solving this problem? The first
method, where the exponent was distributed in can be applied to the first problem
on this page, whereas the second method cannot.
We will now apply the "distribute in" method to the first problem presented
on this page.
(x3y4)5
(x3*5y4*5)
x15y20
This method will also work when the terms are being divided, like the problem below:
(x2 / x)3
Again, the exponent is just "distributed" in:
(x2*3 / x3)
(x6 / x3)
x3
Fractions
Fractions are really just a division problem which is shown in a special form. Since we can just "distribute" in the exponents
for an ordinary division problem, we can do the same for a fraction. Look over the example below:
We can just distribute in the 3, as in the other problems.
As you can see, once the 3 was distributed, the parentheses could be removed. Then the 23 was simplified.
Homework Help: Math: Algebra
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