Hey it’s Professor Dave, I wanna teach you
synthetic division.
We just learned a couple techniques that we
can use to solve polynomials, and in doing
so, we learned that we can find the factors
of polynomials just like we can find the factors
of numbers.
When we do prime factorization, we take some
number and split it up into factors until
we are left with only prime numbers.
That begs the question, are there prime polynomials?
The answer is yes, and if we have a complex
polynomial, it may be prime, meaning it can’t
be factored, or it could be that we can split
it up into a number of different factors.
It may not always be easy to tell whether
a polynomial can be factored.
For example, X to the sixth plus one is prime,
it can’t be factored.
But X to the sixth minus one is not prime,
it can indeed be factored.
So we need a way to test whether a polynomial
is divisible by a particular expression or
not.
That technique is called synthetic division.
It works a lot like long division, which we
learned at the beginning of this course, except
that we are using algebraic expressions instead
of numbers.
This will come in handy if we have something
like a cubic or quartic equation, and we know
one of the zeroes.
Let’s say we are looking at X to the fourth
plus X cubed minus eleven X squared minus
five X plus thirty, and we want to see if
X equals two is one of the solutions.
If it is, then X minus two would be a factor,
and that could allow us to simplify the expression.
The way we will test this is as follows.
Let’s make two lines at a right angle like
this, and in this top row we will list only
the coefficients of our equation.
Then we want to test to see if two is one
of the solutions, so let’s put a two to
the left over here.
Now we take the first coefficient, in this
case a one, and drop it down here.
Now we multiply by the two, and put that here.
Now we can add these together to get three,
and write that here.
Again, we multiply by two to get six.
Adding to negative eleven, we get negative
five.
Then we multiply to get negative ten, add
to get negative fifteen, multiply to get negative
thirty, and finally we add to get zero.
The fact that this is a zero right here means
that there is no remainder from the division,
which means that two is a solution, and therefore
X minus two is indeed a factor of this polynomial.
These resulting numbers here tell us the coefficients
of the polynomial that remains after dividing
by X minus two.
We just have to understand that what is left
will be of a degree that is one less than
the original polynomial, so since we started
with a quartic, the result is a cubic.
We can simplify this to X minus two times
X cubed plus three X squared minus five X
minus fifteen.
If we wanted, we could continue factoring
from there until we have nothing but binomials,
and therefore all the possible solutions.
Say we continue with this one and try to see
if negative three is a solution.
We would perform the same algorithm as before,
dropping the one down, multiplying to get
negative three, adding to get zero, multiplying
to get zero, adding to get negative five,
multiplying to get fifteen, and adding to
get zero.
Once again, with a remainder of zero, we see
that negative three is indeed a solution,
and X plus three is therefore a factor, so
leaving the X minus two as it is, we can factor
this cubic term further to get X plus three
times X squared minus five, which we get from
the coefficients we just calculated.
That last bit is easy to solve, in a separate
equation, we just bring five over here, take
the square root, and we get plus or minus
root five.
That means we have successfully found all
four solutions to this original quartic equation,
and those are two, negative three, root five,
and negative root five.
So this is the best way to solve for equations
of a degree higher than second degree.
You may be wondering, how do we decide which
possible solutions to test for?
That’s a very good question, so let’s
go over the rational roots test.
This is something we can do to get a list
of possible solutions to try.
To get this list, we just look at the expression,
like this one, two X cubed plus three X squared
minus three X minus two, and we make a fraction.
On the numerator of this fraction will go
all the factors of the constant term, which
are one and two.
On the denominator go all the factors of the
coefficient of the leading term, which are
again one and two.
We also put a plus or minus to the left of
this fraction.
To get all the likely zeroes to test, we just
make all the possible fractions.
So we could try plus or minus one, plus or
minus one half, and plus or minus two.
That is still six possibilities to try, but
at least it gives us a short list of good
options.
Remember, these are not all solutions, just
a list of possible solutions.
Sometimes, none of these will end up being
solutions, it’s just a helpful trick that
will usually shorten our search.
Let’s try and see if X equals one is a solution.
We do our synthetic division by setting things
up as we learned before, with the coefficients
two, three, negative three, and negative two.
We place the one here, drop the first two
down, and get to multiplying and adding.
One times two is two, three plus two is five,
one times five is five, negative three plus
five is two, one times two is two, and negative
two plus two is zero.
That means that one is indeed a solution.
Remember, if this last digit was anything
other than zero, then the number you are testing
is not a solution.
You can cross it out, and try something else
from the list.
But this one did work, so we can split up
this cubic function into X minus one, since
one was a solution, and then a quadratic function
given by the coefficients here, which will
be two X squared plus five X plus two.
From there, we could go back to our list of
zeroes and try another.
Perhaps negative two will work.
So we can keep the same workspace, put a negative
two here, drop the two down, and off we go.
Negative two times two is negative four, five
plus negative four is one, negative two times
one is negative two, two plus negative two
is zero.
Once again, the zero means that this is a
solution, so we can break this quadratic up
into X plus two, since negative two is a solution,
and the resulting binomial from synthetic
division, two X plus one.
Looking at this last binomial, it is clear
that if we solve this term by itself, we get
negative one half, and so the three zeroes,
or solutions of this cubic are one, negative
two, and negative one half.
You can imagine how this can get much trickier
for quartics or quintics, especially if the
rational roots test gives us a long list of
possible zeroes to try, but it does give us
a rigorous way to solve for these tricky higher
order polynomials, which is indeed an impressive
feat, so let’s check comprehension.
