The real fun in algebra comes from factoring, it is an essential step in the solving of most equations. We will just scratch the surface here, as this page is aimed at the Algebra 1 student (more will be added later). We will be covering the GCF or Greatest Common Factor (It is also called the Greatest Common Divisor - same thing), for Polynomials, and some of the special products.
When a number or variable divides evenly into another number or variable, it is called a factor. The GCF 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 GFC. I am, of course, partial to my method - so here it is.
There are some very methodical foolproof methods for finding the GCF. 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 GFC - please write me so I know. I could include other methods if anyone thinks they would be helpful.
Now that you have the GCF - what do you do with it? Let's find the GCF 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 GCF as far as the Coefficient goes. Now lets look at the variables, we have
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 GCF 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. So, the total GFC is 3x. We divide each term by the GCF and put it outside the parenthesis and get 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.
At this point if 3x = 0, then the whole equation = 0 because 0 times anything is 0, and 3x is being multiplied times the rest of the equation. The same holds true for the (3x2 - 2x + 2) part of the equation. How do you find x in the equation 3x2 -2x + x = 0? I'm glad you ask, and I'll answer in a bit.
The first step is always finding any GFC's. Once that is done, we can try additional methods. It's at this point that I have to tell you about something that strikes fear in the hearts of many a mathematician. The dreaded Quadratic Formula:
|−b ± √ b2 − 4ac|
One sure fire way to factor any quadrilateral is to use the Quadratic Formula. It works every time, but it can be kind of a pain. We get the a, b, and c in the quadratic formula from the a, b, and c in the standard quadratic equation which is ax2 + bx + c = 0.
Another method is checking the coefficients. We want to end up in the form (x + a)(x + b) you could substitute either of those pluses with a minus sign. Checking the Coefficients works best when the Coefficient of the squared variable is 1. If it's not 1, we need to look at a different method. Let's do an example x2 + 3x + 2 = 0.
We need to look for two numbers that multiplied together equal the last term and add together to equal the coefficient of the middle term. In this case 2*1 = 2, the last term and 2+1 = 3 the middle term. So, (x+2)(x+1) = 0. Now, we set each equal to 0 and we get
x+2 = 0, subtract 2 from both sides and end up with x=-2
x+1 = 0, subtract 1 from both sides and end up with x=-1
It's a little harder when a ≠ 1 in a quadratic equation. Remember, we always have the quadratic formula to fall back on. Anyway, this is the formula: ax2 + bx + c = 0 And we are dealing with a situation where the value of "a" is not 1. How do we turn this formula into something like (fx + g)(hx + j)? We have to satisfy three conditions -
a = f*h
c = g*j
b = f*j + g*h
Let's work with an example -
So, a = 6, b = 19, and c = 10. There is no GCF. So, we need to look at the factors of a and c to see if they can be combined to equal b.
a = 6 = 6*1 and c=10 = 10*1, 6*1 + 1*10 = 16, so that's not it.
a = 6 = 1*6 and c = 10 = 10*1, 1*1 + 6*10 = 61, but we need c = 19.
a = 6 = 2*3 and c = 10 = 10*1, 2*1 + 3*10 = 32, no.
a = 2*3 and c = 2*5, 2*5 + 3*2 = 16 again - not it.
a = 3*2 and c = 2*5, 3*5 + 2*2 = 19 !!! That's it!!!
So, in (fx + g)(hx + j)
a = f*h = 3*2 = 6
c = g*j = 2*5 = 10
b = f*j + g*h = 3*5 + 2*2 = 15 + 10
f = 3, g = 2, h = 2, and j = 5
-6x2 + 19x + 10 = 0
(3x + 2)(2x + 5) = 0
Well, we can't have a bunch of negative a's running about. We don't have a problem with any of the coefficients being negative, with the exception of a. If a is negative, count it as a part of the GCF. For instance, if our formula was --6x2 + 19x + 10 = 0 factor out the negative and we get -(6x2 - 19x - 10 = 0 ). Now, solve the problem, remember that minus sign is out there and re-multiply times any root values you get.
At some point, if you are going to be factoring, you will need to memorize (I know - dirty word), the special products. I have given these their own special products page. I do this because it is an important topic, and because this page is getting a bit long.
XIII. Linear Equations
XVII. Real Numbers
XIV. Math Terms
XXI. Pythagorean Theorem
XXVI. Special Products
XXVII.. Square Root Functions
XXVIII. Systems of Equations
XXIX, Variable Mathematics