Professor Dave here, let’s learn how to FOIL.
We’ve learned how to add and subtract polynomials, so now it’s time to learn how to multiply
them.
But before we do this with long expressions, let’s start with the tiny ones.
First we can multiply monomials.
Two X times three is six X, just the way two apples time three is six apples.
Two X times three X is six X squared.
In this case, the numbers multiply, as do the X’s, so that’s easy enough as well.
Now we have to look at binomials.
These are expressions with two terms.
If we have a monomial times a binomial, again this is very simple, because we just apply
the distributive property.
Looking at two X times the quantity X minus four, we just multiply across, so we get two
X squared minus eight X.
But if we want to multiply two binomials, things start to get a little less obvious.
Take X plus three times X minus two.
What we want to understand is that in order to perform this calculation, we would have
to distribute each term in one binomial across the other term, essentially using the distributive
property twice, and a quick way to do this is to apply the FOIL method.
These letters stand for first, outer, inner, last, and they instruct us as to which numbers
to multiply together and in what order.
First means we multiply the first term in each binomial.
In this case X times X equals X squared.
Then we do the outer terms, X times negative two, or negative two X.
Then the inner terms, three times X, or three X, and finally we do the last terms in each
binomial, or three times negative two, which is negative six.
That gives us X squared minus two X plus three X minus six, and we can combine like terms
to get X squared plus X minus six.
Again, the reason this works is because it’s as though we are distributing each term of
one binomial across the other binomial, and the FOIL method just ensures that we do it
the same way every time so that we don’t make a mistake.
The most common error when foiling is certainly listing the wrong sign for a particular term.
Don’t get confused by the plus or minus in between the terms of the binomial, just
think of the binomial as a list of two terms, each with a particular sign.
When multiplying terms, don’t think about whether the terms in the resulting list will
be added or subtracted, just list the result with its sign, whether positive or negative,
and the operations will take care of themselves.
For example, when we did the X times two, just assume that this minus sign means negative
two, and when we get negative two X, we just place it in the list.
For three times X, we got positive three X, so just list positive three X, with a plus
symbol.
At the end, all the operations are there for us to follow, clear as day, so if you use
this method it will be hard to make a mistake.
Let’s try a trickier one, (X squared plus four) times (negative two X minus one).
These binomials look different than the previous example, but we still just FOIL as usual.
For the first term, we get X squared times negative two X, which is negative two X cubed.
For the outer, since there’s a negative one, we get negative X squared.
Then inner, we get negative eight X, and lastly four times negative one is negative four.
There are no like terms to combine, so that’s the end of it.
Now that we can multiply binomials, what happens when we multiply a binomial by a trinomial?
As one might expect, it’s a little more complicated, but it’s essentially the same
thing.
We could multiply the first term of the binomial across the trinomial, and then the other.
For example, X plus two Y times three X minus Y minus four.
First let’s multiply X across the trinomial.
That gives us three X squared minus XY minus four X.
Then we just continue this list of terms by multiplying two Y across the trinomial.
That will be six XY minus two Y squared minus eight Y.
There are two like terms to combine, the XY terms.
Six XY minus one XY gives five XY.
Then we just list according to convention, X squared then Y squared, then XY, then X,
then Y, and there we have it.
So it’s not really any harder as the expressions get bigger, just more steps, since we need
to use the distributive property more times.
If we wanted to multiply a trinomial by a trinomial, it wouldn’t be that hard, we
would just end up with nine terms, and then condense from there if any like terms exist.
Let’s check comprehension.