- #1
Ackbach
Gold Member
MHB
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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:
Examples of the Method
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.
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.
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.
- $a_j=0$. Then $\min(a_j,s_j)+\min(a_j,t_j)=0+0=0=a_j$, 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.
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