# AC Method for Factoring Trinomials

• MHB
• Ackbach
In summary, the method for factoring a trinomial into a product of two dissimilar binomials involves writing the trinomial as the product of two binomials and examining the factor pairs of the product of the leading coefficient and constant term to find the correct factor pair. The method also includes dividing out the greatest common factor of the coefficients in each binomial. This method has been proven to be correct and results in a unique factorization, assuming that the quadratic can be factored.

#### Ackbach

Gold Member
MHB
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 \\
&=\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.
In all cases, $\min(a_j,s_j)+\min(a_j,t_j)=a_j$, showing that the method, if it's possible to execute, does indeed result in a final expression that is
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.

Last edited by a moderator:
Thanks @Ackbach! What forum do you think we could move this to?

Greg Bernhardt said:
Thanks @Ackbach! What forum do you think we could move this to?
STEM Educators and Teachers again, I think.

## What is the AC Method for Factoring Trinomials?

The AC Method is a technique used to factor trinomials by splitting the middle term into two terms whose sum is equal to the middle term and whose product is equal to the product of the first and last terms.

## How do I use the AC Method to factor a trinomial?

To use the AC Method, first write the trinomial in the form ax^2 + bx + c. Then find two numbers, a and c, whose product is equal to the product of a and c. These numbers must also add up to b. Once you have found these numbers, rewrite the middle term of the trinomial as a sum of these two numbers. Finally, factor the resulting expression by grouping and factor out the common factor.

## What are the benefits of using the AC Method for factoring trinomials?

The AC Method is a straightforward and organized approach to factoring trinomials. It can be especially useful when the leading coefficient of the trinomial is not 1 or when the trinomial has a large number of terms. Additionally, the AC Method can be used to factor trinomials that cannot be factored using other methods.

## Are there any limitations to using the AC Method for factoring trinomials?

While the AC Method is a powerful tool for factoring trinomials, it can only be used for trinomials with a leading coefficient of 1. It also may not work for trinomials with certain patterns, such as difference of squares or perfect squares.

## Can the AC Method be used for factoring other types of polynomials?

Yes, the AC Method can also be used to factor other types of polynomials, such as quadratics or higher degree polynomials. However, the process may be more complex and require additional steps compared to factoring trinomials.