Example 3 Factor
each of the following polynomials.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
(f) [Solution]
(g) [Solution]
Solution
(a)
Okay since the first term is x^{2} we know that the factoring must take the form.
We know that it will take this form because when we
multiply the two linear terms the first term must be x^{2} and the only way to get that to show up is to
multiply x by x. Therefore, the first
term in each factor must be an x. To finish this we just need to determine
the two numbers that need to go in the blank spots.
We can narrow down the possibilities considerably. Upon multiplying the two factors out these
two numbers will need to multiply out to get 15. In other words these two numbers must be factors
of 15. Here are all the possible ways
to factor 15 using only integers.
Now, we can just plug these in one after another and
multiply out until we get the correct pair.
However, there is another trick that we can use here to help us
out. The correct pair of numbers must
add to get the coefficient of the x
term. So, in this case the third pair
of factors will add to “+2” and so that is the pair we are after.
Here is the factored form of the polynomial.
Again, we can always check that we got the correct answer
my doing a quick multiplication.
Note that the method we used here will only work if the
coefficient of the x^{2}
term is one. If it is anything else
this won’t work and we really will be back to trial and error to get the
correct factoring form.
[Return to Problems]
(b)
Let’s write down the initial form again,
Now, we need two numbers that multiply to get 24 and add
to get 10. It looks like 6 and 4
will do the trick and so the factored form of this polynomial is,
[Return to Problems]
(c)
Again, let’s start with the initial form,
This time we need two numbers that multiply to get 9 and
add to get 6. In this case 3 and 3
will be the correct pair of numbers.
Don’t forget that the two numbers can be the same number on occasion
as they are here.
Here is the factored form for this polynomial.
Note as well that we further simplified the factoring to
acknowledge that it is a perfect square.
You should always do this when it happens.
[Return to Problems]
(d)
Once again, here is the initial form,
Okay, this time we need two numbers that multiply to get 1
and add to get 5. There aren’t two
integers that will do this and so this quadratic doesn’t factor.
This will happen on occasion so don’t get excited about it
when it does.
[Return to Problems]
(e)
Okay, we no longer have a coefficient of 1 on the x^{2} term. However we can still make a guess as to the
initial form of the factoring. Since
the coefficient of the x^{2}
term is a 3 and there are only two positive factors of 3 there is really only
one possibility for the initial form of the factoring.
Since the only way to get a 3x^{2} is to multiply a 3x and an x these must
be the first two terms. However,
finding the numbers for the two blanks will not be as easy as the previous
examples. We will need to start off
with all the factors of 8.
At this point the only option is to pick a pair plug them
in and see what happens when we multiply the terms out. Let’s start with the fourth pair. Let’s plug the numbers in and see what we
get.
Well the first and last terms are correct, but then they
should be since we’ve picked numbers to make sure those work out
correctly. However, since the middle
term isn’t correct this isn’t the correct factoring of the polynomial.
That doesn’t mean that we guessed wrong however. With the previous parts of this example it
didn’t matter which blank got which number.
This time it does. Let’s flip
the order and see what we get.
So, we got it. We
did guess correctly the first time we just put them into the wrong spot.
So, in these problems don’t forget to check both places
for each pair to see if either will work.
[Return to Problems]
(f)
Again the coefficient of the x^{2} term has only two positive factors so we’ve only
got one possible initial form.
Next we need all the factors of 6. Here they are.
Don’t forget the negative factors. They are often the ones that we want. In fact, upon noticing that the coefficient
of the x is negative we can be
assured that we will need one of the two pairs of negative factors since that
will be the only way we will get
negative coefficient there.
With some trial and error we can get that the factoring of this
polynomial is,
[Return to Problems]
(g)
In this final step we’ve got a harder problem here. The coefficient of the x^{2} term now has more than one pair of positive
factors. This means that the initial
form must be one of the following possibilities.
To fill in the blanks we will need all the factors of
6. Here they are,
With some trial and error we can find that the correct
factoring of this polynomial is,
Note as well that in the trial and error phase we need to
make sure and plug each pair into both possible forms and in both possible
orderings to correctly determine if it is the correct pair of factors or not.
We can actually go one more step here and factor a 2 out
of the second term if we’d like to.
This gives,
This is important because we could also have factored this
as,
which, on the surface, appears to be different from the
first form given above. However, in
this case we can factor a 2 out of the first term to get,
This is exactly what we got the first time and so we
really do have the same factored form of this polynomial.
[Return to Problems]
