# Factoring non-monic quadratics

I understand how factoring a quadratic $(ax^2 + bx +c)$ with "$a$" equal to one works. If it's in the form $(x+d)(x+e)$ then "$d$" and "$e$" are the only two terms that can multiply together to yield a term with no "$x$". And when "$d$" is multiplied by the 2nd "$x$", and "$e$" is multiplied by the first "$x$" and the like terms are added together they yield "$bx$". Thus, with this information we can get the factored form.

Question: How does the "trick" that math teachers teach us to factor quadratics with "$a$" not equal to 1 work? This is the trick where we figure out what multiplies to "$ac$" and what adds to "$bx$", then rewrite the quadratic in the form $ax^2 + dx + kx + c$ and factor from there.

Note: I've been using this trick for a long time, and this isn't my first time with quadratics, but this is a question that's been bugging me for a while. I'm sorry if the answer is glaringly simple. Thanks for your help.

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• I don't remember the trick that you're talking about. Could you show us a worked example of it so that we clearly know what you're talking about? – Tanner Swett Aug 21 '18 at 17:39
• @TannerSwett Solve $x^2+5x+6=0.$ We look for two numbers $r,s$ whose sum is $-5$ and whose product is $6.$ Those are the roots of the equation. And we can factor $x^2+5x+6=(x-r)(x-s).$ – mfl Aug 21 '18 at 17:47

## 4 Answers

For $a$ not equal to $1$, begin like this $$a x^2 + bx + c = a \left(x^2 + \frac{b}{a} x + \frac{c}{a}\right)$$ then factor the part inside the parentheses using the method of $a=1$.

Example $$5x^2-7x+2 = 5\left(x^2 - \frac{7}{5}x + \frac{2}{5}\right) =5\left(x-1\right)\left(x-\frac{2}{5}\right)$$

I don't know what you mean by a "trick" but an obvious thing to do when you have $ax^2+ bx+ c$ with $a\ne 1$ is to factor the "a" out! That gives you $a(x^2+ \frac{b}{a}x+ \frac{c}{a})$. Now you can use whatever "tricks" you like to factor $x^2+ \frac{b}{a}x+ \frac{c}{a}$.

It is the same "trick". Note that we have

$$ax^2+bx+c=0\iff x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0.$$ So the roots $r,s$ satisfy

\begin{cases} r+s&=-\dfrac ba\\ rs &=\dfrac ca \end{cases}

When $a=1$ we get

\begin{cases} r+s&=-b\\ rs &= c \end{cases}

Consider the equivalent polynomial (equivalent in terms of factorization, anyway) with leading coefficient $1$, namely, $x^2+(b/a)x+(c/a)$. You can break this into two factors, $x+r_1$ and $x+r_2$. Multiplying these two factors together will give you

$$(x+r_1)(x+r_2)=x^2+(r_1+r_2)x+r_1r_2$$

So we have two ways of writing the same polynomial, one in terms of its coefficients and one in terms of its factors. Equating the two, we see that

$$x^2+(r_1+r_2)x+r_1r_2=x^2+(b/a)x+(c/a)$$

So it's clear that the numbers $r_1$ and $r_2$ must have a sum of $b/a$ and a product of $c/a$. The original polynomial $ax^2+bx+c$ will then be factored into $a(x+r_1)(x+r_2)$, where you are free to distribute the leading coefficient $a$ however you wish between the two factors.