# What are some methods to quickly factorize quadratic equations? For example, turning `x^2-4x+4` into `(x-2)^2` in a couple of seconds, or `2x^2-5x-3` into `(x-3) (2x+1).`

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### 1 Answer

There are several methods and variations of these basic methods for factoring quadratics. If we start from a product of the form `(x+a)(x+b)` and expand it, we get

`(x+a)(x+b)=x^2+(a+b)x+ab.` **(1)**

Notice that the coefficient of `x`, which is `a+b,` is the *sum* of the two numbers `a` and `b` and the constant term is the *product* of `a` and `b.` Thus, considering your example of `x^2-4x+4,` if we can find two numbers `a` and `b` that add to `-4` and multiply to `4,` we can work from right to left in equation **(1)**. In fact, we can set `a=-2,` `b=-2,` since `-2+(-2)=-4` and `(-2)(-2)=4.`

This proves that `x^2-4x+4=(x-2)(x-2).`

Your second example is a little more complicated because the coefficient of `x^2` is no longer `1.` However, we can do something similar.

`(ax+b)(cx+d)=(ac)x^2+(ad+bc)x+bd.` **(2)**

Here, notice that the coefficient of `x` , which is `ad+bc,` is the sum of two numbers `ad` and `bc,` while the product of the coefficients of `x^2` and the constant term (this product being `adbc)` is equal to the product of these two numbers. So in your example of `2x^2-5x-3,` we look for two numbers that multiply to `-3*2=-6` and add to `-5.` A little trial and error shows that `-6` and `1` are two such numbers. Now break up the `x` term into these two numbers as follows:

`2x^2-5x-3=2x^2-6x+x-3.` Now we can factor by grouping. I placed parentheses to make it clearer

`(2x^2-6x)+(x-3)=2x(x-3)+(x-3)`

`=(2x+1)(x-3).`

Like I said, there are variations of these methods. One **guaranteed method that will never require trial and error** **and work every time** is to just use the quadratic formula. If we use the quadratic formula on

`2x^2-5x-3,` we see that the zeros are `-1/2` and `3,` so `(x+1/2)` and `(x-3)` are factors of this quadratic. The leading coefficient is `2,` so we put a `2` in front and get

`2x^2-5x-3=2(x+1/2)(x-3)=(2x+1)(x-3).`

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