Professor Dave here, let’s learn how to factor polynomials.
We just learned how to multiply binomials by the FOIL method, and understanding this
technique will allow us to express simple polynomials in the form AX squared plus BX
plus C equals zero.
This is the standard form for a quadratic, or second degree polynomial, with A, B, and
C being some integers.
Remember that when we solve something like this, what we are asking is: what value of
X will make this equation true?
In other words, what number can I plug in for X such that this expression will indeed
be equal to zero?
The first technique we will use to solve some simple trinomials will be the reverse of the
FOIL method we just learned, in that FOILing two binomials gives us a polynomial, whereas
we will now take a polynomial and express it as the product of two binomials.
To do this we have to be a little clever and think backwards.
First we have to understand that when we multiply binomials, like X plus M times X plus N, we
get X squared plus MX plus NX plus MN.
We can combine the X terms and get M plus N times X.
So what that means, is that if we have some polynomial in the form of AX squared plus
BX plus C, where A equals one, in order to find out which binomials could multiply together
to get this polynomial, we need to find the two numbers, M and N, that go in the binomials,
and they must be numbers that multiply together to give C, and add together to give B. Let’s
do an example and I’ll show you what I mean.
Take X squared plus seven X plus ten.
Because the first term is X squared, the binomials can both start with X, so we just have to
figure out which numbers go in these two spots.
The two numbers must multiply to give ten and add to give seven, so let’s start by
listing pairs of factors for ten.
In this case there are very few options, plus or minus ten and one, and plus or minus five
and two.
Remember, the product of two negatives is a positive, so negative factors are possible
here.
Now we ask ourselves, which of these pairs of factors can add or subtract in some way
to give seven?
Ten and one won’t work, since we could get eleven or nine, either positive or negative,
depending on the signs that go in the binomials, but we can’t get seven.
Five and two will work, however, since five plus two is seven.
So that means we can put positive five here and positive two here, and we have successfully
factored the polynomial.
We can check our work by multiplying through.
X squared plus two X plus five X plus ten, after combining like terms, does indeed give
us the original expression.
Notice that we add five and two to get the coefficient of seven X, and we multiply five
and two to get ten, which is precisely what we were doing in reverse by selecting five
and two.
Let’s try another.
X squared minus eight X minus twenty.
Just like before, we always start by listing the pairs of possible factors for the C term.
Things get a little trickier when the last term is negative, since we have one and negative
twenty, but also negative one and positive twenty.
Then there are both combinations of two and ten, as well as five and four, so we have
six pairs of factors that might work.
But once again, though any of these pairs will multiply to give negative twenty, only
one of them will add to give negative eight.
We go through the options, trying the math in our head.
Five and four is no good, nor is one and twenty, but with ten and two, we could have negative
ten and positive two, and that does add to negative eight.
So we can write X minus ten times X plus two, and that’s the answer, which we can then
FOIL if we want to check the answer.
Let’s do one more that is a little trickier.
Three X squared plus five X minus two.
Now that there is a coefficient of three by the X squared, that means one of the binomials
has to start with three X, and the other with X, because that’s the only way to get three
X squared for the first term.
Now that also means that it is not arbitrary which other number goes where, because one
will multiply by three X and the other by X.
Nevertheless, let’s get our factors and see what happens.
We have negative two, so luckily there are only two options, it’s either positive one
and negative two, or negative one and positive two.
So we have to ask ourselves, which pair of factors will allow us to multiply one factor
by three X and the other by X, and add those together to get five X.
First we try one and negative two.
That could give us three X minus two X, but that’s no good.
X minus six X also doesn’t work.
How about negative one and positive two.
We quickly see that if we multiply the two by three X and the negative one by X, we get
six X minus X, or five X.
So we place the negative one and the two into these binomials making sure to put them in
the correct spots.
We need the two to multiply the three X to get six X, so we have to place it in the other
binomial so that they do indeed multiply.
We do not put it in the same binomial as three X.
And there we have it, three X minus one times X plus two, which once again, we can FOIL
to double check our answer.
At this point, you may be wondering, why are we doing this?
How does this help us solve polynomials?
Well the answer is simple.
If any of these polynomials had been set equal to zero, then the binomials must multiply
to give zero.
And we know that anything times zero is zero, so finding a value for X that will make either
of these binomials equal to zero will necessarily satisfy the equation.
For example, with our first attempt, we factored this expression to get X plus two times X
plus five.
If that had been set equal to zero, we can now look at the first term.
How do we get X plus two to be equal to zero?
Well that’s easy, if X plus two equals zero, then X must be negative two, because negative
two plus two is zero.
That means that if X is negative two, this first term is zero, and then it doesn’t
matter what this other term is, because zero times anything is zero.
Likewise, if X is negative five, this term is zero, and the whole thing is zero.
That means that negative two and negative five are the two solutions to this quadratic,
which we found simply by factoring the expression.
The second one we did gave us X minus ten times X plus two.
By similar reasoning, either positive ten or negative two can be plugged in for X to
make this equation valid, so those are the two solutions.
For the last one, we ended up with three X minus one times X plus two.
Of course negative two works, but for this other one we just need one extra step.
Let’s make a separate equation, where three X minus one equals zero.
Add one to both sides, divide by three, and we get X equals one third, so those are the
two solutions in that case.
For any of these examples, we can always check our answers by plugging them into the original
polynomial to verify that it does indeed equal zero, which is great for peace of mind on
a test.
And with that, we have learned a simple technique to solve polynomials.
Not all polynomials can be solved by this method, as we will later see, but lots of
them will, so let’s check comprehension.