XXV.  Factoring Polynomials

If you wish to solve for the variable, you will need to start by factoring polynomials.  The first step in factoring is to find the G.C.F.  (Greatest Common Factor).

Polynomials are expressions with variables that have an exponent greater than or equal to 2.  Or, for a more complete definition check out Dictionary.com.

A colorful graph depicting nothing, really.  But doing it in very bright colors of blue, yellow and red.

A Factoring Polynomials with Greatest Common Factor (GCF)

When a number or variable divides evenly into another number or variable, it is called a factor.  The G.C.F.  is the largest number or variable that will divide evenly into two or more terms.  I have read a lot of different methods for finding the G.F.C..  I am, of course, partial to my method - so here it is.

There are some very methodical foolproof methods for finding the G.C.F of polynomials.  The Tree and the Ladder Method to name just two.  But there's really no mystery, so I'm going to leave them out of this discussion.  If you've read this and are confused as to how to find the G.F.C. - please write me so I know.  I could include other methods if anyone thinks they would be helpful.

The general method for finding the G.C.F. applies regardless of the exponents involved.  Now that you have the G.C.F. - what do you do with it?  Let's find the G.C.F. of this equation:  9x3 - 6x2 +6x = 0.  The smallest coefficient is 6, but 6 won't divide evenly into 9, so 6 won't work.  But 6 = 3*2.  We know 2 won't go into 9, but 3 will go evenly into all three coefficients.  So 3 is our G.C.F. as far as the Coefficient goes.  Now lets look at the variables, we have x3, x2, and x, x is common to all 3 terms.  Our equation becomes 3x(3x2 - 2x + 2) = 0  after we divide each term by 3x.  Putting it outside the parenthesis like that, means we can multiply 3x times each term if we want, to get the original equation.  Remember, when a variable has an exponent it can have up to that exponents value in answers.  So, we could have up to 3 values for x that satisfy this equation.  If we were trying to solve the equation, we now have two of the factors and could break it up into two equations.

3x = 0 and (3x2 - 2x + 2) = 0.  From the first equation we see that one value must be x = 0.  We can solve the other factor and get all three values of x.  The easiest way to solve the second equation is using the quadratic formula.  The formula is based on ax2 + bx + c = 0, so in this case a = 3, b = -2, and c = 2.

The quadratic formula gives us -b plus or minus  √(b2 - 4ac)/2a  now we plug and play

-(-2) +/- ((-2)2 - 4(3)(2))/2(3) = 2 +/- (4 - (24))/6 - We are on the verge of ending up with (-20) in our formula.  There is no real number that can be multiplied times itself to equal -20.  So, this equation only has one real root, and that is 0.  We do get to work with imaginary numbers - but, not at this juncture.

Factoring Quadrilaterals

Rather than just copying the web page on Factoring Quadrilaterals, why not just click here - Factoring.  Factoring quadrilaterals is a special situation of factoring polynomials, quadrilaterals have a factor of 2.   I will, however, discuss factoring polynomials with higher degrees.   Again, I apologize for copying some of the information included in Special Products.  But, here goes - 

SP #3 The Sum of Two Cubes

Cubes or variables with a power of three, are an instance of factoring polynomials that comes in two flavors.  The sum and the difference of two cubes.  Sure, there's other instances - but they offer more difficulties.

x3 - a3 = (x + a)(x2 - ax + a2) is one of the special products that deals with perfect cubes.  An example might be 8x3 + 27, which is the same as (2x)3 + 33, so it would equal (2x + 3)((2x)2 - (2x)(3) + 32) = (2x + 3)(4x2 - 6x + 9) 

SP #4 The Difference of Two Cubes

x3 - a3 = (x + a)(x2 + ax + a2).  An example of this would look like 8x3 - 27 = (2x - 3)(4x2 + 6x + 9)

Factoring Polynomials by Grouping

24x3 - 18x2 + 40x - 30 = 0.  To factor this we use grouping.  Group the first two terms together and factor that, then the second two terms and factor that.  Looking at 24x3 - 18x2 I see that I could factor out a 6 and and x2 from each term.  This leaves me with 6x2(4x - 3), use the distributive property to multiply that out and you're back at the original first two terms.

The second two terms 40x - 30 are both evenly divisible by 10, giving us 10(4x - 3).  Put the two factored terms back together and we have:

6x2(4x - 3) + 10(4x - 3) = 0.  Now, I have two terms and each of them can be evenly divided by (4x-3).  This would give me (4x-3)(6x2 + 10) = 0.  If either of these factors - 0 then the whole equation would be equal to zero, so we solve.

First the 4x - 3 = 0

4x = 3

x = 3/4

Now the 6x2 + 10 = 0

6x2 = -10

x2 = -10/6 = -5/3

x = (-5/3) and, once again, we end up with an imaginary number because we can't have the square root of a negative number.

What to do with all the other polynomials?

Factoring polynomials with higher exponents can be a bit tricky.  Of course there are equations that just don't fit the scenarios we've covered.  You're goal is to manipulate them into quadratics or lower and then either factor them or use the quadratic formula.

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