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The functions you are most familiar with are probably polynomial functions.
What are polynomial functions?
A polynomial function in the variable is a function which can be written in the
form where the ’s are all constants (called the coefficients) and is a whole
number (called the degree when ). The domain of a polynomial function is
Which of the following are polynomial functions?
The phrase above “in the variable ” can actually change. is a polynomial in , and is
a polynomial in .
Multiplying polynomials is based on the familiar property of
arithmetic, distribution: .
Multiply by .
Multiply by .
We’ll start by distributing the first polynomial into the
second. Then we’ll distribute back, and finish by combining like terms.
After combining like-terms, we find
The result is that we have multiplied every term of the first polynomial by
every term of the second, then added the results together. In the case of two
binomials (polynomials with only two terms), this is frequently referred to as
There are several product formulas that arise repeatedly when working with binomials.
You will likely have seen most of these before.
If is the polynomial function given by , find . Find .
means to replace in the
formula for by .
In the same way, asks us to replace by .
If , find .
Factoring is a bit like an inverse operation of multiplying polynomials. We start
with the multiplied out polynomial, and ask what the individual factors
The easiest factors to deal with are common factors.
The coefficients of this
polynomial have a greatest common divisor of , and the least degree of the terms is ,
so is a common factor. Taking this out leaves: .
For trinomials (polynomials with three terms) of the form , we try to factor as .
We’ll start with an example with .
Since the leading coefficient is , we want
to factor as . If we multiply out we get . We need to find two numbers that add to ,
and multiply to . Look at the different ways of multiplying two numbers to get : , , ,
... That last one, and are two numbers that add to and multiply to give
Our factorization is .
The process is slightly more complicated when .
In this case, want to find .
Ignore that middle term for now, and just focus on the leading term and the constant
term. We need to find numbers that multiply to for our coefficients, and that
multiply to for the constants in each factor. That means we’re looking at things like
, , , . Actually, that’s it, as can only factor as or , and only factors as
All of these possibilities will give the right term and the right constant term. The
only difference is the -term they give. Do any of them give an -coefficient of ?
Remember, a root of a polynomial function is an -value where the polynomial is zero.
There is a close relationship between roots of a polynomial and factors of that
If is a root of a polynomial, then is a factor of that
Notice that if we substitute into the polynomial, we get zero out. That
means is a factor of . Polynomial long division gives:
The quotient was , so . It remains to factor the quotient.
The leading coefficient factors as , and factors as and . Examining the
possibilities, we find the factorization of as . The entire polynomial factors as
A quadratic equation in is an equation which is equivalent to one with the form ,
where , , and are constants, with .
There are three major techniques you are probably familiar with for solving quadratic
Completing the Square.
Each of these methods are important. Factoring is vital, because it is a valid
approach to solve nearly any type of equation. Completing the Square is a technique
that becomes useful when we need to rewrite certain types of expressions. The
Quadratic Formula will always work, but has some limitations.
Solve the quadratic equation
We’ll start by writing this in it’s standard form: .
Notice how there is a common factor of 2? Let’s divide by 2.
. Any of the above methods will work here, so let’s try factoring. What numbers add
to and multiply to ? and do. That gives us:
Either (giving us ) or (giving us ). The two solutions are .
Solve the quadratic equation
Again, we’ll write it in standard form: . The quadratic
does not factor, so we’ll complete the square instead.
Start by moving the constant term to the other side This has . That means and .
Add to both sides. The left-hand side is now a perfect square, . We can then solve
using square roots.
If you had used the quadratic formula, , instead of factoring or completing the
square above, you would have found the same solutions.
Solve the quadratic equation .
none of the above
Building on these methods, we’ll consider more general types of equations. A
polynomial equation is an equation which is equivalent to one of the form .
Start by rewriting as . This is not a quadratic equation in , but it IS a
quadratic in . We continue by completing the square.
Either , or . Since is negative, only is possible.
The only real solutions are .
Factoring is a process that helps us solve more than just quadratic equations, as long
as we first get one side of the equation equal to zero.
Solve the equation
way for a product of real numbers to be zero, is for one of the factors to have been
zero. Here, that means either , or or . This gives us three equations, each
substantially less complicated than the original one, to solve. By solving them, we
Start by rewriting the equation as . This is not a quadratic-type equation, so
we resort to factoring.
Thus, either (so ), or (so ), or (so ).
The solutions are .
The next theorem above is a deep fact of mathematics. The great mathematician
Gauss proved the theorem in 1799.
The Fundamental Theorem of Algebra Every
polynomial of the form where the ’s are real (or even complex!) numbers and has
exactly (possibly repeated) complex roots.
Recall that the multiplicity of a root indicates how many times that particular root
is repeated. The Fundamental Theorem of Algebra tells us that a polynomial
equation of degree , will have exactly complex solutions, once multiplicity is taken
into account. As non-real solutions appear in complex-conjugate pairs (if our
equation has real coefficients), we will always have an even number of non-real
solutions. That means, taking multiplicity into account, a quadratic equation will
have either or real solutions. A cubic equation will have either or real
The equation has as one solution. Find another solution.
Notice that there were only two real numbers that solve this equation. How is that
possible, if we said a cubic equation has to have either or real solutions?