Tip:
Highlight text to annotate it
X
♪Music♪
Welcome to Georgia Highlands College Math97 in Math99 video
tutorials. In this video segment will be answering the question
how do you factor in trinomial whose leading coefficient
is not 1 using the trial and error method? Well, the first
step in this method is to set up what you know your factors are
going to look like including those Xs that multiply
together to give you that X^2 term. Secondly your going to
find all factor pairs of A and then all factor pairs of C.
You're going to then try various combinations of the factor
pairs from A and C that are going to add up to make B.
which is your middle term. And finally, you'll check your
answer using multiplication. Let's look at an example.
Here we have 5X^2-14X+8 and the first step in the trial and
error method tells us to set up parentheses that
we know that we will factor into. So that's what I've done
over here is given myself a room to write the factors that
I find.
The second step is to find all the factor pairs of A. Well in
this particular example A is 5 and since 5 is prime and
actually only has one factor pair which is 1 and 5 or -1 and
-5. Our third step is to find factor pairs of 8.
We have 1 and 8 and we also have 2 and 4. So we found
all of the factor pair of A. And all of the factor pairs of C.
Now the next step in this process, this is why it's called
trial and error, is you're just going to start trying various
combinations of these factor pairs to see which set
adds up to that middle term -14X.
I'll start with (5X+2)(X+4) So I'm going to
show you where that came from. I got the 5 first from here
and my 2 came from this factor here. So that's that first set
of parentheses. The second set has an understood 1 as its
factors, you don't see a 1 sitting in front of that X, and
then I have the 4 sitting there. So I want to try all the
combinations using these factors before I move on to the
different set of factors being the 1 and the 8 in the
second slots of each set of parentheses. So I'm just going
to start mixing and matching all the possibilities so we can
have (5X+4) and (X+2). so I just flip-flopped the last
two terms in each set of the parentheses there.
Or I could have (5X-2) and (X-4).
or I could have (5X-4) and (X-2). I haven't tried
combinations where these signs are different because
I know the signs need to be the same because
those last two numbers have to multiply together to make 8.
So I either need to be multiplying together a negative
and a negative or positive and a positive so I have
not mixed matched my signs and done one positive and
one negative, that would give me -8 at the end.
So now we just come back and start multiplying
these out and see which ones and it's really the center
term we're concerned with because we know every time we
multiply 5X by X we get 5X^2 which is our first term
and we noted every time we multiply 2 and 4 that we always
get 8. So it's just that middle term that we need to be
concerned with and we get the middle term by multiplying the
outside term then the inside terms. So for this first one we
have 20X plus 2X which gives us 22X which is a no go
because were trying to equal out to -14X. So moving along we'll
go ahead and try multiplying our next set. See what that middle
term comes out to. Here we have 10X+4X which gives us 14X.
Wel it's really close but it's the wrong sign so we still
need to keep on going to multiplying the outside terms
and the inside terms for this particular one we get 5X times
-4 which is -20X and -2 times X is -2X which gives us
a total of -22X, still not there. And last but by far not
least, multiplying the outside and the inside terms for this
last set of binomial factors we get -10X and -4X
of course it would be the last one, gives us a -14X. So through
trial and error we've find that (5X-4)(X-2) gives us those
inside terms. So we can come up here and fill in 5X and a -4
positive 1, which is understood, and a -2.
So we need to check this to make sure that every saying works out
as it should by simply multiplying our binomial factors
And we do that with distribution of each term in the first
binomial with each term in the second binomial so 5X times X is
5X^2, 5X times -2 is -10X. -4 times X gives -4X, and -4 times
-2 is 8. And when I combine my like terms
I get 5X^2-14X+8. +8 . So since our original
polynomial matches are checked polynomial we have that
(5X-4)(X-2) is the factored form of that trinomial.
I hope this has answered your questions about the method of
trial and error for factoring trinomials.
If you still have questions, please make sure to
contact your Highlands instructor. Thank you.
♪Music♪