Solving Quadratic Equations ( Factorization 2 )

A Method For Simple Cases

Luckily there is a method that works in simple cases.
For an equation

Step 1: Find two numbers that add to give b, and multiply to give ac

Example:  
2x² + 7x + 3

b is 7, and ac is 2×3 = 6
So we want two numbers that add up to 7, and multiply together to make 6.
In fact 6 and 1 do that (6+1=7, and 6×1=6)

Step 2:

Rewrite the middle with those numbers.

Rewrite 7x with 6x and 1x:

2x² + 6x + x + 3

Step 3: Factor the first two and last two terms separately:

The first two terms 2x² + 6x factor into 2x(x+3)
The last two terms x+3 don't actually change in this case
So we get:

2x(x+3) + (x+3)

Step 4: If you've done this correctly, your two new terms should have a clearly visible common factor.

In this case you can see that (x+3) is common to both terms

So we can now rewrite it like this:

2x(x+3) + (x+3) = (2x+1)(x+3)

Check: (2x+1)(x+3) = 2x² + 6x + x + 3 = 2x² + 7x + 3 (Yes)

Much better than guessing!

Example 2 :
6x² + 5x - 6

Step 1:
b is 5, and ac is 6×(-6) = -36

One of the numbers has to be negative to make -36, so by playing with a few different numbers I find that -4 and 9 work nicely:

-4+9 = 5, and -4×9 = -36

Step 2: Rewrite 5x with -4x and 9x:

6x² - 4x + 9x - 6

Step 3: Factor first two and last two:

2x(3x - 2) + 3(3x -2)

Step 4: Common Factor is (3x - 2):

(2x+3)(3x - 2)

Check: (2x+3)(3x - 2) = 6x² - 4x + 9x - 6 = 6x² + 5x - 6 (Yes)