how do you factorise a quadratic where there is a number in front of the x squared?

All quadratic equations have the form ax² + bx + c.

If you take the equation: 2x² + 4x + 2

In this example a=2, b=3 and c=2

In order to factorise what we call a number (coefficient) higher than 1 in front of the x² we use something called the ac method. 

so lets try to factorise 3x² - 2x -5

step 1. Take the a term and the c term and multiply them together (short form ac). So in this example, since a = 3 and c = -5,   a*c= -15

step 2.  next we try to find 2 numbers which multiply to become ac = -15 but add up to become the b term, in this case -2

so -15 has a number of possible solutions. 

-53, -35, -151, -115

out of these only -5 and 3 add up to become -2.

step 3. Now that we have found the numbers which multiply to become -15 and add up to become -2. We input these into brackets like a normal quadratic.

(3x-5)(3x+3) 

However this isnt the final step as if you expand these brackets you will not reach 3x²-2x-5

step 4. Divide the brackets we have reached by the number in front of the x² - the a term

so for (3x-5)(3x+3) 

we look at which bracket is divisible by the 3. Obviously the bracket (3x+3) is easily divisible by 3. So once you divide that by 3, you reach (x+1) 

Now the answer reads:

(3x-5)(x+1)

this is now our final answer. 

It may seem difficult and confusing at first but since it is only 4 steps it quickly becomes second nature.