Math 151 - ICA #4 Summer 2016 Section 8349
Instructor: John Kwon
In-Class Assignment #4: Factoring by Grouping - Solution
• Factor each polynomial completely.
1.
5x(x − 4) − 2(4 − x)
Answer. Note that the red-colored (x − 4) and (4 − x) are almost the same expressions, except
that they have opposite signs from each other. This means that (4 − x) = −(x − 4), so −2(4 − x) =
+ 2(x − 4). Then the given polynomial becomes
5x(x − 4) − 2(4 − x) = 5x(x − 4) + 2(x − 4) .
Factoring out the common factor now yields
5x(x − 4) + 2(x − 4) = (x − 4)(5x + 2) .
2.
(m2 − 3mn + 3m) − (2m − 6n + 6)
Answer. Note that m is the GCF of the expression inside the first set of parentheses, and 2
is the GCF of the expression inside the second. Factoring out these GCFs from the respective
expressions gives
(m2 − 3mn + 3m) − (2m − 6n + 6) = m(m − 3n + 3) − 2(m − 3n + 3) .
Now we can see that (m − 3n + 3) is the common factor. Factoring this out yields
m(m − 3n + 3) − 2(m − 3n + 3) = (m − 2)(m − 3n + 3) .
3.
a3 − 7 + 7a2 − a
Answer. There are four terms, so we need to try the method of factoring by 2-by-2 grouping,
except the polynomial needs to be rearranged so that factoring by grouping can be visualized
clearly. In fact, rearranging the terms in the descending order of powers works out well:
a3 −7 + 7a2 −a = a3 + 7a2 −a − 7
Note now that a2 can be factored out of the first two terms after the rearrangement, and −1 can
be factored out of the last two terms. Doing this yields
a3 + 7a2 −a − 7 = a2 (a + 7) − (a + 7) .
Factoring out the common factor now yields:
a2 (a + 7) − (a + 7) = (a + 7)(a2 − 1) .
Finally, observe that (a2 − 1) is the difference of two squares, and therefore can be factored as
(a2 − 1) = (a + 1)(a − 1) . Therefore we get
(a + 7)(a2 − 1) = (a + 7)(a + 1)(a − 1) .
4.
t6 − t4 −t2 + 1
Answer. Once again, there are four terms, so we should factor by grouping 2-by-2. This time
the polynomial can be grouped without rearranging the terms. Factoring out t4 from the first two
terms and −1 from the last two gives
t6 − t4 −t2 + 1 = t4 (t2 − 1) − (t2 − 1) .
Factoring out the common factor yields
t4 (t2 − 1) − (t2 − 1) = (t2 − 1)(t4 − 1) .
Note that (t2 − 1) is a difference of two squares and can be factored as (t2 − 1) = (t + 1)(t − 1) ,
so we have
(t2 − 1)(t4 − 1) = (t + 1)(t − 1)(t4 − 1) .
Also, note that (t4 − 1), too, is a difference of two squares (t2 and 1), so we can factor it as
(t4 − 1) = (t2 + 1)(t2 − 1), yielding
(t + 1)(t − 1)(t4 − 1) = (t + 1)(t − 1)(t2 + 1)(t2 − 1) .
Now, observe that (t2 − 1) is again a difference of two squares, with (t2 − 1) = (t + 1)(t − 1), so
we get
(t + 1)(t − 1)(t2 + 1)(t2 − 1) = (t + 1)(t − 1)(t2 + 1)(t + 1)(t − 1) .
Finally, since the factors (t + 1) and (t − 1) appear twice, we write them as powers:
(t + 1)(t − 1)(t2 + 1)(t + 1)(t − 1) = (t + 1)2 (t − 1)2 (t2 + 1) .