Factoring a quadratic equation means expressing it in two brackets. A quadratic equation is one in the form ax2+bx+c = 0.

### Factoring when a = 1

Firstly, make sure the equation is in the standard format, as above. Then, to make things easier for you, write down the two brackets -

( x )( x ) = 0.

Now, you need to find two numbers that multiply to give c, and add/subtract to give the coefficient of b. Then you simply put these numbers into the brackets, and sort out the +/- signs correctly.

Example:
x2 + x = 12

First, we rearrange into the standard format-

x2 - x - 12 = 0

We need two pairs of numbers hat multiply to give 12 and add/subtract to give 1. The numbers are clearly 3 and 4.

Now, write the equation -

(x + 3)(x - 4) = 0

Then, as an essential check, we'd expand the brackets.

### Factoring when a does not equal 1

Such as:

3x 2 + 7x + 6 = 0

The method is the same is pretty much the same as before. In this case we'd have to find two numbers that multiply with the 3x and x terms in the brackets and then add or subtract to give the value of b. So in this case it'd be 2 in the first bracket and 3 in the second.

Although Noung's writeup shows a good way for a human to factor a trinomial, trying to program an algorithm using that method would be horribly inefficient. It would have to first factor a and c, and then test every possible combination of factors until it finds the correct one. Luckily, there is another method that is much more efficient:

ax²+bx+c

as before, but first split the x term into 2 parts, so that the product of those two coefficients equals the product of a and c. An easy way to do this would be to iterate through the integers between either ±sqrt(ac) or ±b, whichever is greater, since this will cover all possible values. If none fits this, its factorization is irrational or imaginary, and another method will be necessary to factor it. If a number is found, however, the result will look like this:

(ax²+b1x)+(b2x+c)

where b1b2=ac. Next, factor out everything possible from both sets of parentheses. The binomials left inside each set of parentheses will be the same:

px(rx+s)+q(rx+s)

where pr=a, ps=b1, qr=b2, and qs=c. Finally, the distributive property gives you

(px+q)(rx+s)

###### Example:
6x²+5x-4

First, find 2 numbers with a sum of 5 and a product of -24. They happen to be 8 and -3. So split the x term up like this:

(6x²+8x)+(-3x-4)

Then, factor a 2x out of the first binomial, and a -1 out of the second:

2x(3x+4)-1(3x+4)

Now, use the distributive property to factor it:

(2x-1)(3x+4)

Using this method, you don't need to factor any numbers, as you do in the normal method, and the amount of numbers you need to test to find the correct ones is also much smaller.

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