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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:

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

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

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

$ax^2+bx+c=0$

user16795
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    The Vieta formulas in degree $2$? – K B Dave May 17 '19 at 04:38
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    "The ac method" is essentially the same, you find two factors of $ac$ that add $b$. – Julian Mejia May 17 '19 at 04:45
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    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. – Servaes May 17 '19 at 05:00
  • @Servaes Sorry, I think I fixed that? – user16795 May 17 '19 at 12:43
  • Ah okay, thank you all! AC method & Vieta's formula is what I was looking for. If anyone wants to post an answer, I'll give y'all the points. :) – user16795 May 17 '19 at 12:51
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    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. – Bill Dubuque May 17 '19 at 14:02

1 Answers1

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This is method is known to me as middle term factor.

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 $(x-\alpha)(x-\beta)$.

Vineet
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