p.
140. #33.
Next, notice that this is a SUM of two cubes!
The FIRST is
,
which is actually
and
the SECOND is
which
can be written
.
Remember that the sum of two cubes factors into the
product of a binomial times a trinomial in this form, according to the
formula
.
What you have is:



p.
141. #37.
The first step in any factoring problem should be to take out the common
factor. In this case the common factor is



Next, notice that this is a SUM of two cubes! The FIRST is
,
which is actually
and
the SECOND is
which
can be written
.
Remember that
the sum of two cubes factors into the product of a binomial times a
trinomial in this form, according to the formula
.
What you have is:



p. 142. # 51.
Solution:
This is a grouping
problem in which it works to group the first two terms together and the
second two term together. From the first two terms you can take out a
common factor of
.
From the second two terms, there really isn’t a common factor to take out,
so just factor out a
.

Now there is a common factor of
,
so take out the common factor:

Next, factor the
difference of two squares:

And again, the
difference of two squares:

p. 142. # 52.
Solution:
You may have guessed from the
grouping by parentheses and by the colors used in this problem, that it is a
trinomial. Factor the first grouping as a trinomial, then factor the middle
two terms by taking out a common factor of 4, and leave the last 4 alone:


And even better, you can write it
like this:

p. 142. # 56.
Solution:
This is a grouping
problem in which it works to group the first two terms together and the
second two term together. From the first two terms you can take out a
common factor of
.
From the second two terms, there really isn’t a common factor to take out,
so just factor out a
.

Now there is a common factor of
,
so take out the common factor:

Next, factor the
difference of two squares:

And again, the
difference of two squares:
