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- Jan 26, 2012

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**Description of the Method**

We are given a trinomial of the form $ax^{2}+bx+c$, and asked to factor it into a product of two dissimilar binomials $(fx+u)(gx+v)$. The method that follows assumes $a,b,c$ have no common factor; if they do, you must factor out the greatest common factor before proceeding.

The method is as follows:

- Write $ax^{2}+bx+c$ as $(ax+\underline{\phantom{45}}\,)(ax+\underline{\phantom{45}}\,)$.
- Examine the factor pairs of the product $ac$, and see which pair, when added together, make $b$. Call this pair $s,t$. You now write $(ax+s)(ax+t)$.
- For each binomial, divide out the greatest common factor of the coefficients. That is, for $ax+s$, divide out the greatest common factor of $a$ and $s$, and for $ax+t$, divide out the greatest common factor of $a$ and $t$.
- The result is $(fx+u)(gx+v)$, where

\begin{align*}

f&=\frac{a}{\text{gcf}(a,s)} \\

u&=\frac{s}{\text{gcf}(a,s)} \\

g&=\frac{a}{\text{gcf}(a,t)} \\

v&=\frac{t}{\text{gcf}(a,t)}.

\end{align*}

**Examples of the Method**

- Factor $15x^{2}+29x-14$. There is no gcf, so we examine the product $15\times (-14)=-210$. The pair products of $-210$ that add to $29$ are $35$ and $-6$. Hence, we write

$$15x^{2}+29x-14=\left(\frac{15x+35}{5}\right)\left(\frac{15x-6}{3}\right)=(3x+7)(5x-2).$$ - Factor $12x^{2}-46x-36$. This one has a greatest common factor, so we get that out of the way as $2(6x^{2}-23x-18)$. The pair products of $6(-18)=-108$ that add to $-23$ are $-27$ and $4$. So we write

$$2(6x^{2}-23x-18)=2\left(\frac{6x-27}{3}\right)\left(\frac{6x+4}{2}\right)=2(2x-9)(3x+2).$$

Note that you can use the prime factorization of $ac$ to find all factor pairs, and hence the correct factor pair. Also note that if $ac<0$, you are looking for the difference between the two factors of the factor pair, and if $ac>0$, you are looking for the sum.

**Proof of the Method**

We assume that it is possible to execute the method. If the method does not execute, then I claim the quadratic does not factor. First, we show that multiplying out the result yields the original quadratic. That is,

\begin{align*}

(fx+u)(gx+v)&=fgx^2+fvx+ugx+uv \\

&=fgx^2+(fv+ug)x+uv \\

&=\left(\frac{a}{\text{gcf}(a,s)}\cdot \frac{a}{\text{gcf}(a,t)}\right)x^2

+\left(\frac{a}{\text{gcf}(a,s)}\cdot\frac{t}{\text{gcf}(a,t)}+

\frac{s}{\text{gcf}(a,s)}\cdot \frac{a}{\text{gcf}(a,t)} \right)x \\

&\qquad+\frac{s}{\text{gcf}(a,s)}\cdot \frac{t}{\text{gcf}(a,t)} \\

&=\left(\frac{a^2}{\text{gcf}(a,s)\cdot \text{gcf}(a,t)}\right)x^2

+\left(\frac{a(s+t)}{\text{gcf}(a,s)\cdot\text{gcf}(a,t)}\right)x

+\frac{st}{\text{gcf}(a,s)\cdot\text{gcf}(a,t)} \\

&=\left(\frac{a^2}{\text{gcf}(a,s)\cdot \text{gcf}(a,t)}\right)x^2

+\left(\frac{ab}{\text{gcf}(a,s)\cdot\text{gcf}(a,t)}\right)x

+\frac{ac}{\text{gcf}(a,s)\cdot\text{gcf}(a,t)} \\

&=\frac{a}{\text{gcf}(a,s)\cdot\text{gcf}(a,t)} \, (ax^2+bx+c).

\end{align*}

Recall that $st=ac$, and $s+t=b$. So, for this method to work, we must show that

$$\frac{a}{\text{gcf}(a,s)\cdot\text{gcf}(a,t)}=1.$$

By the Fundamental Theorem of Arithmetic, we can write

\begin{align*}

a&=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k} \\

b&=p_1^{b_1}\cdot p_2^{b_2}\cdots p_k^{b_k} \\

c&=p_1^{c_1}\cdot p_2^{c_2}\cdots p_k^{c_k},

\end{align*}

where the $p_j$ are primes, and $a_j, b_j$ and $c_j$ are non-negative integers. Note that we have included all primes in either $a$ or $b$ or $c$'s factorization, and recall that $\text{gcf}(a,b,c)=1$. This forces $\min(a_j,b_j,c_j)=0$ for all $j$. Next, suppose that

\begin{align*}

s&=p_1^{s_1}\cdot p_2^{s_2}\cdots p_k^{s_k} \\

t&=p_1^{t_1}\cdot p_2^{t_2}\cdots p_k^{t_k},

\end{align*}

are the prime factorizations of $s$ and $t$. Because $ac=st$, it must be that $a_j+c_j=s_j+t_j.$ Then

\begin{align*}

\text{gcf}(a,s)&=p_1^{\min(a_1,s_1)}\cdot p_2^{\min(a_2,s_2)} \cdots

p_k^{\min(a_k,s_k)} \\

\text{gcf}(a,t)&=p_1^{\min(a_1,t_1)}\cdot p_2^{\min(a_2,t_2)}

\cdots p_k^{\min(a_k,t_k)}.

\end{align*}

We are attempting to prove that

$$a=\text{gcf}(a,s)\cdot\text{gcf}(a,t),$$

or, equivalently, that

$$a_j=\min(a_j,s_j)+\min(a_j,t_j)$$

for all $j$. This equation is certainly not true in general; however, I argue that it is true in our special case here.

We examine the equation $s+t=b$. Let $z=\text{gcf}(s,t)$. Then $z|b$. But if $z>1$, then $z$ cannot divide both $a$ and $c$, or else $\text{gcf}(a,b,c)=z>1$. Then

$$\text{gcf}(a,b,c)=1\implies \text{gcf}(a,s+t,c)=1\implies

\text{gcf}(a,c,s,t)=1\implies \min(a_j,c_j,s_j,t_j)=0.$$

We break this down by cases.

- Suppose $c_j=0$. Then $a_j=s_j+t_j$, implying that $a_j>s_j$ and $a_j>t_j$. Hence, $\min(a_j,s_j)+\min(a_j,t_j)=s_j+t_j=a_j,$ as required.
- Suppose $c_j>0$. Then we have two subcases:
- $a_j=0$. Then $\min(a_j,s_j)+\min(a_j,t_j)=0+0=0=a_j$, as required.
- $a_j>0$. Then either $s_j$ or $t_j$ is zero. Without Loss of Generality, we may assume $s_j=0$. Then $a_j+c_j=t_j$, implying $a_j<t_j$. It follows that $\min(a_j,s_j)+\min(a_j,t_j)=0+a_j=a_j$, as required.

equivalent to the original expression.

By the Fundamental Theorem of Algebra, if the quadratic factors, it factors uniquely. Since we have found a factorization, assuming it has worked, we have found the correct factorization.

Questions and concerns can go in the commentary thread.