Answer: How do you factor out a quadratic equation?

Here is the base equation:
ax²+bx+c


Here is the equation I will be using for this tutorial:
x²-13x+42

With this equation...
a = 1 (default- if no specific "a" value is given, it is automatically 1)
b = -13
c = 42

The first thing you have to do is determine the factors of c (42).

1 * 42
2 * 21
3 * 14
6 * 7

Remember, when you multiply 2 negatives together, you still get a positive.

-1 * -42
-2 * -21
-3 * -14
-6 * -7

So, all those factors have potential.  How do we determine which set to use?  Simply determine which pair of numbers will ADD up to the value of "b" (which, in this case, is -13)

1 + 42 = 43
2 + 21 = 23
3 + 14 = 17
6 + 7 = 13

-1 + -42 = -43
-2 + -21 = -23
-3 + -14 = -17
-6 + -7 = -13     <--- This is the right one

We have determined that the numbers -6 and -7 give a product of 42 and a sum of -13.  That was the hard part.  All you have to do now is plug it in.  For this example, we'll just equate -6 to "d" and -7 to "e."  You don't necessarily have to have the numbers in that order- they can be switched.  (i.e. you could make -6 "e" and -7 "d," etc.)  Use this formula:

(x + d)(x + e)

The "x" is in there because that's the variable that is in this function.  If you plug in "d" and "e," it should look something like this:

(x + -6)(x + -7) or, when simplified so you're not adding negatives, (x - 6)(x - 7)

And that's your final answer: (x - 6)(x - 7)

To check your answer, simply re-expand it.  (If this is being read by someone who doesn't know quadratics very well, expanding means working out the new problem you have created)

Using the equation you've created: (x -6)(x - 7) multiply each variable / number in the first set of parentheses by each variable / number in the second set of parentheses.  Example:

x * x = x²

x * -7 = -7x  
-6 * x = -6x
-6 * -7 = 42


Then, add your answers together: x²  + -7x + -6x + 42.
This equals your original equation of:  x² - 13x + 42
If this result is not the same as your original equation, you have done something wrong along the way and need to start over.



What if "a" is more/less than 1 (default)?

Good question.

Remember, the base equation is ax² + bx + c

Here's an example function:  2x² + 20x + 50

In this example, the current value of 'a' is 2.

First, restructure the equation by relocating the 2 to the outside of the rest of the function.  You can achieve this by "dividing" the equation by 2.  Then, to make sure that this new restructured equation is equivalent to the original, put the 2 on the outside with your new equation in parentheses.  It should look like this:

2(x² + 10x + 25)

Then, you can use the previously discussed method to factor out the equation that's in the parentheses.  5 * 5 = 25, and 5 + 5 = 10.  Therefore, you would use 5 for both "d" and "e" in your factored form.

2(x + 5)(x + 5)




What if there is no "b" value?

There's always a "b" value- in this case, it is simply equal to zero.  Take this equation for example:

2x² - 32

Remember the "divide by 'a'" trick?

2(x² - 16)

Then factor it out.  Remember, 16 is a square number.

2(x - 4)(x + 4)

Those 4's aren't included in the original function because -4 + 4 = 0 (which is the "b" value).  However, -4 * 4 = -16 (the "c" value), so that was included.

Any questions?  CONSTRUCTIVE criticism?  Comment!