# Name for this method of factoring quadratic and are there any textbooks that describe it?

I remember learning this method of factoring quadratics in middle school or high school, but looking for a name or more information on it leads me to dead ends.

Given:

$$ax^2+bx+c=0$$

$$d*e=a*c$$

$$d+e=b$$

Then the factorization of the quadratic is:

$$(x+\frac{e}{a})*(x+\frac{d}{a})$$

Proof:

$$(x+\frac{e}{a})*(x+\frac{d}{a})=0$$

$$x^2+\frac{ex+dx}{a}+\frac{ed}{a^2}=0$$

$$x^2+\frac{x(e+d)}{a}+\frac{ed}{a^2}=0$$

Via substitution of the given above:

$$x^2+\frac{bx}{a}+\frac{ac}{a^2}=0$$

$$x^2+\frac{bx}{a}+\frac{c}{a}=0$$

$$a*(x^2+\frac{bx}{a}+\frac{c}{a})=a*(0)$$

$$ax^2+bx+c=0$$

• The Vieta formulas in degree $2$? Commented May 17, 2019 at 4:38
• "The ac method" is essentially the same, you find two factors of $ac$ that add $b$. Commented May 17, 2019 at 4:45
• Your 'proof' is very confusing. You write down a bunch of equations out of the blue that don't seem to hold necessarily, without any explanation. And some are plain false, such as $$x^2+\frac{bx}{a}+\frac{ac}{a^2}=ax^2+bx+c=0,$$ as well as the next equality. Commented May 17, 2019 at 5:00
• @Servaes Sorry, I think I fixed that? Commented May 17, 2019 at 12:43
• This is often called the AC method. In the linked post I describe how it works for any degree polynomial. You can find almost 30 worked examples in the linked questions on that thread. Commented May 17, 2019 at 14:02

Lets take an example $$f(x)=x^2+6x+8$$. We have to find two numbers such that their sum is their product is $$8$$ and the sum is $$6$$. So, factors are $$(x+4)(x+2)$$.
In general, $$ax^2+bx-c$$ here constant term $$ac$$ is negative so we have to find two numbers such that their difference is $$b$$ and the product is $$ac$$.
Sometimes finding what to add or subtract might be difficult in that case we can use quadratic formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$. You will get two solutions from here $$x=\alpha,\beta$$. Hence your required factors will be $$(x-\alpha)(x-\beta)$$.