Factoring Polynomials
Before we actually get to factoring a full polynomial there are a couple of things we need to know. The first thing is prime
factors of numbers. Before we talk about how to find the prime factors of numbers we need to know what a prime
number is.
~Prime Numbers: are integers in which there are no factors other than itself and 1. In other words, the only way to
multiply and get the number is by one and itself.
~Examples of Prime Numbers:
2 <-- only way to get two by multiplication is with 1 and 2.
3 <-- only way to get three by multiplication is with 1 and 3.
5
7
etc.
~Examples of Non-Prime Numbers:
4 <-- the factors of four are 1 and 4 (1*4=4) and 2 (2*2=4)
6 <-- factors of six are 1 and 6 (1*6=6), 2 and 3 (2*3=6)
8 <-- factors of eight are 1 and 8 (1*8=8), 2 and 4 (2*4=8)
etc.
Now that we know what Prime Numbers are we can continue on with Prime factorization.
~Prime Factorization: is the method of breaking down non-prime numbers into a group of prime numbers. The idea of
this is to take really large numbers and rewrite them as multiplication between smaller numbers.
There are two ways of showing this. The first way is dividing the number you want to factorize by
prime numbers (usually by 2 if the number is even). The second way is by splitting the number
into two smaller numbers and continuing this pattern until you get to all prime numbers.
~Example of Division by Prime:
96 ÷ 2 = 48
48 ÷ 2 = 24
24 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 3 = 1
96 = 2 * 2 * 2 * 2 * 2 * 3
~Example of Factor Tree (number splitting):
As you can see, it does not matter which method you prefer to use the answer should be the same no matter what
method is used. Next we will discuss factoring Monomials.
~Prime Factorization of a Monomial: is the method of breaking down a monomial into a group of prime
numbers along with variables. In other words, you break the
monomial down into a multiplication between prime numbers and
variables.
~Example of Prime Factorization of a Monomial:
Factor: 48 * x^(3) * y^(2)
48 ÷ 2 = 24
24 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 3 = 1
48 = 2 * 2 * 2 * 2 * 3
X^(3)÷ X = X^(2)
X^(2) ÷X = X
X ÷ X = 1
X^(3) = X * X * X
Y^(2) ÷ Y = Y
Y ÷ Y = 1
Y^(2) = Y * Y
So that 48 * X^(3) * Y^(2) = 2 * 2 * 2 * 2 * 3 * X * X * X * Y * Y.
Now you can also rewrite this as 48*X^(3)*Y^(2) = 2^(4) * 3 * X^(3) * Y^(2)
Now that we can factor a term of a polynomial (monomials are a one term polynomial if you don't remember) we can start
finding the Greatest Common Factor (GCF) of a polynomial.
~Greatest Common Factor (GCF): is the largest factor that all the terms in a given polynomial have.
~Examples of GCF:
Find the GCF of 14*X^(4) + 7*X*Ythen show the factored form of the polynomial
Factors of 14*X^(4) Factors of 7*X*Y
14 ÷ 2 = 7 7 ÷ 7 = 1
7 ÷ 7 = 1 7 * 1 = 7
14 = 7 * 2 X ÷ X = 1
X^(4) ÷ X = X^(3) X * 1 = X
X^(3) ÷ X = X^(2) Y ÷ Y = 1
X^(2) ÷ X = X Y * 1 = Y
X ÷ X = 1
X * X * X * X = X^(4)
14*X^(4) = 2 * 7 * X * X * X * X 7*X*Y = 7*X*Y
(it was already in its prime state)
Final Answer: GCF := 7 * X
Factored Form: 14*X^(4) + 7*X*Y = 7*X*(2*X^(3)) + 7*X*(Y) = 7*X*(2*x^(3) + Y)
Now we will factor Quadratic Trinomials in standard form.
~Factoring Quadratic Trinomials: The reason we want to factor these quadratic trinomials is we want to simplify
the quadratic that is given to us. Writing polynomials in their factored forms can
prove to be helpful in many situations. There are a lot of different ways to do this.
The method shown here is the standard way to factor. When factoring quadratic
trinomials you should end up with two binomials being multiplied together.
Factor: x^(2) + 2*x + 1
First we need to remember how we can build one of these before we can factor them. Lets look at
multiplying two binomials together. Take (x+1)*(x+1) , using the foil method, (x+1)*(x+1)=x^(2) + x + x + 1 =
= x^(2) + 2*x+1 which happens to be the polynomial we are trying to factor now. Now, notice how the middle
term and last term are formed. These two are very important because they are tied together in how their signs
form (the leading term is always positive in standard form which means the factors of the leading term will
always be positive). Notice how the last term, usually a constant, is only formed by multiplication while the
middle term is formed with addition. Here we go....
Step One: Factor leading term: x^(2)
x * x = x^(2) this tells us that each binomial will have a leading term of x.
Step Two: Factor out Last Term: 1
1 * 1 = 1 this tells us that each binomial will have a last term of 1, but we do not know the
sign of either of them nor do we know which binomial to put them in.
Step Three: Figure out the signs of your factors: + or -?
This is where logic comes into play. Look at the sign of the last term in the trinomial. It happens to
be positive in this case. There are two ways in which you can get a positive sign by multiplication
those are either positive times a positive or a negative times a negative. Now look at the second
term of the trinomial. The sign is positive as well. Which means we need to use two positive signs
in our factored out binomials.
Step Four: Figure out which set of factors add to the middle coefficient: 2
1+1=2 So now we know that 1 and 1 are the constants we need
So if we put all the clues from steps one to four together we get our answer of: (x + 1) * (x + 1)
Step Five: Check answer:
(x + 1) * (x + 1) = x^(2) + x + x + 1 = x^(2) + 2*x + 1
Factor: x^(2) - 4*x + 3
Step One: Factor Leading Term: x^(2)
x * x = x^(2) so that each binomial will start with x
Step Two:Factor Last Term: 3
3 * 1 = 3
Step Three: Figure out the signs
First looking at the last term we see that it is positive. This tells us the signs in the binomials
must either be positive or negative. Next look at the term in the middle, since it is negative the
signs used in our answer must be negative as well.
Step Four:Figure out which set of factors add to the middle coefficient: - 4
- 3 - 1 = - 4
Now add the clues to get the answer: (x - 3) * (x - 1)
Step Five: Check Answer:
(x - 3) * (x - 1) = x^(2) - 3*x - x + 3 = x^(2) - 4*x + 3
The last thing we will look at is Factor by Grouping.
~Factor by Grouping: is a tool used on polynomials with more than three terms. Note that this can be used
on polynomials with only three terms as well, but it is mainly used on four terms and
above. The idea is to get the terms that are most alike together and then to "pull out" the
common factors between the two terms (in other words find their GCF).
Factor: x^(3) +2* x^(2) + 3*x + 6
Step One: Collect the terms with the most things in common with each other.
If you look at x^(3) and 2*x^(2) you can see that these two have much more in common than
say x^(3) and 3*x. You can see this by finding their respective GCF's. Which leaves you to
group 3*x with 6.
Step Two: Find and factor out the GCF of each "group".
GCF of x^(3) and 2*x^(2) is x^(2)
GCF of 3*x and 6 is 3
We now rewrite the given polynomial as...
[x^(3) + 2*x^(2)] + [3*x + 6] = x^(2) * [x + 2] + 3 * [x + 2]
group 1 group 2
Step three: Simplify if possible.
If you look closely you can see that we can factor something out of both groups. Both groups
have x + 2 in them so if we factor them out we get....
x^(2) * [x + 2] + 3 * [x + 2] = (x + 2) * (x^(2) + 3) which is our final answer for this.
That was a quick look at polynomials and factoring. Thank you for reading.
If you have any questions email me at: [email protected]