Recall: One way factor, say, $x^2+3x+2$, is to write $$x^2+3x+2=(x+\mbox{___})(x+\mbox{___})$$ and fill in the numbers by trial and error.
We can do something similar for trinomials of the form $ax^2+bx+c$.
Example: $15r^2-28r+12$
We write our factorization as $$\color{magenta}{15}r^2-28r\color{blue}{+12}=(\color{magenta}{\mbox{___}}r+\color{blue}{\mbox{___}})(\color{magenta}{\mbox{___}}r+\color{blue}{\mbox{___}})$$ The values on the left must multiply to $\color{magenta}{15}$ and the values on the right must multiply to $\color{blue}{12}.$
When we multiply it all out, the middle term should simplify to $-28r.$
We're going to try different combinations of numbers that work and check the result until we get it. $$ \begin{array}{lll} \displaystyle (\color{magenta}{\mbox{3}}r\color{blue}{-\mbox{3}})(\color{magenta}{\mbox{5}}r\color{blue}{-\mbox{4}})&\displaystyle= \color{magenta}{15}r^2-27r\color{blue}{+12} &\mbox{nope}\\ \displaystyle (\color{magenta}{\mbox{3}}r\color{blue}{-\mbox{4}})(\color{magenta}{\mbox{5}}r\color{blue}{-\mbox{3}})&\displaystyle= \color{magenta}{15}r^2-29r\color{blue}{+12} &\mbox{nope}\\ \displaystyle (\color{magenta}{\mbox{3}}r\color{blue}{-\mbox{2}})(\color{magenta}{\mbox{5}}r\color{blue}{-\mbox{6}})&\displaystyle= \color{magenta}{15}r^2-28r\color{blue}{+12} &\mbox{Yes!}\\ \end{array} $$ It took three guesses and checks to find the right factorization $$ 15r^2-28r+12=(3r-2)(5r-6) $$
Other Examples: Factor the following expressions.
$10t^2â19tâ12$
$$
\begin{array}{lll}
\displaystyle (\color{magenta}{\mbox{2}}t\color{blue}{-\mbox{3}})(\color{magenta}{\mbox{5}}t\color{blue}{+\mbox{4}})&\displaystyle= \color{magenta}{10}t^2-7t\color{blue}{-12} &\mbox{nope}\\
\displaystyle (\color{magenta}{\mbox{2}}t\color{blue}{-\mbox{4}})(\color{magenta}{\mbox{5}}t\color{blue}{+\mbox{3}})&\displaystyle= \color{magenta}{10}t^2-14t\color{blue}{-12} &\mbox{nope}\\
\displaystyle (\color{magenta}{\mbox{2}}t\color{blue}{-\mbox{2}})(\color{magenta}{\mbox{5}}t\color{blue}{+\mbox{6}})&\displaystyle= \color{magenta}{10}t^2+2t\color{blue}{-12} &\mbox{nope}\\
\displaystyle (\color{magenta}{\mbox{2}}t\color{blue}{-\mbox{6}})(\color{magenta}{\mbox{5}}t\color{blue}{+\mbox{2}})&\displaystyle= \color{magenta}{10}t^2-26t\color{blue}{-12} &\mbox{nope}\\
\displaystyle (\color{magenta}{\mbox{2}}t\color{blue}{-\mbox{1}})(\color{magenta}{\mbox{5}}t\color{blue}{+\mbox{12}})&\displaystyle= \color{magenta}{10}t^2+19t\color{blue}{-12} &\mbox{Almost!}\\
\displaystyle (\color{magenta}{\mbox{2}}t\color{blue}{+\mbox{1}})(\color{magenta}{\mbox{5}}t\color{blue}{-\mbox{12}})&\displaystyle= \color{magenta}{10}t^2-19t\color{blue}{-12} &\mbox{Yes! (switched signs)}\\
\end{array}
$$
So,
$$
10t^2â19tâ12=(2t+1)(5t-12)
$$
$-24q^2+38q-15$
We first factor out a negative,
$$-24q^2+38q-15=-(24q^2-38q+15)$$
and focus on the expression $24q^2-38q+15.$
$$
\begin{array}{lll}
\displaystyle (\color{magenta}{\mbox{3}}q\color{blue}{-\mbox{3}})(\color{magenta}{\mbox{8}}q\color{blue}{-\mbox{5}})&\displaystyle= \color{magenta}{24}q^2-39q\color{blue}{+15} &\mbox{nope}\\
\displaystyle (\color{magenta}{\mbox{3}}q\color{blue}{-\mbox{5}})(\color{magenta}{\mbox{8}}q\color{blue}{-\mbox{3}})&\displaystyle= \color{magenta}{24}q^2-49q\color{blue}{+15} &\mbox{nope}\\
\displaystyle (\color{magenta}{\mbox{4}}q\color{blue}{-\mbox{5}})(\color{magenta}{\mbox{6}}q\color{blue}{-\mbox{3}})&\displaystyle= \color{magenta}{24}q^2-42q\color{blue}{+15} &\mbox{nope}\\
\displaystyle (\color{magenta}{\mbox{6}}q\color{blue}{-\mbox{5}})(\color{magenta}{\mbox{4}}q\color{blue}{-\mbox{3}})&\displaystyle= \color{magenta}{24}q^2-18q\color{blue}{+15} &\mbox{Yes!}\\
\end{array}
$$
So,
$$
\begin{array}{lll}
\displaystyle -24q^2+38q-15&\displaystyle=-(24q^2-38q+15) &\mbox{}\\
\displaystyle &\displaystyle=-(6q-5)(4q-3) &\mbox{}\\
\end{array}
$$
$24n^2-38n x+15x^2$
$$
\begin{array}{lll}
\displaystyle (\color{magenta}{\mbox{6}}n\color{blue}{-\mbox{5}}x)(\color{magenta}{\mbox{4}}n\color{blue}{-\mbox{3}}x)&\displaystyle= \color{magenta}{24}n^2-38nx\color{blue}{+15}x^2 &\mbox{Yes!}\\
\end{array}
$$
So,
$$24n^2-38n x+15x^2=(6n-5x)(4n-3x)$$
If you don't like the Guess 'n' Check method, you might like the following, more systematic, method.
A More Systematic Method: The $ac$ Method
To factor $ax^2+bx+c$, find two numbers whose sum is $b$ and whose product is $ac$.
Example: Factor $8z^2+37z-15$
A good first step is to make a table of factors. $$ \begin{array}{c|c} ac=8(-15)=-120 & \mbox{sum}\\ \hline -1 \cdot 120 & 119\\ -2 \cdot 60 & 58\\ -3 \cdot 40 & 37 \, \checkmark \\ \end{array} $$ Once we get our factors, we break up the middle term of our trinomial having these numbers as coefficients and factor by grouping. That is, $$ \begin{array}{lll} \displaystyle 8z^2+37z-15 &\displaystyle=8z^2+40z-3z-15 &\mbox{break up middle term in to factors from table}\\ \displaystyle &\displaystyle= 8z(z+5)-3(z+5)&\mbox{factor by grouping}\\ \displaystyle &\displaystyle= (8z-3)(z+5)&\mbox{pretty slick, eh?}\\ \end{array} $$
More Examples
$15x^2+2x-8$
$$
\begin{array}{c|c}
ac=15(-8)=-120 & \mbox{sum}\\ \hline
-1 \cdot 120 & 119\\
-2 \cdot 60 & 58\\
-3 \cdot 40 & 37 \\
-4 \cdot 30 & 26 \\
-5 \cdot 24 & 19 \\
-6 \cdot 20 & 14 \\
-8 \cdot 15 & 7 \\
-10 \cdot 12 & 2 \, \checkmark \\
\end{array}
$$
$$
\begin{array}{lll}
\displaystyle 15x^2+2x-8 &\displaystyle=15x^2-10x+12x-8&\mbox{break up middle term in to factors from table}\\
\displaystyle &\displaystyle= 5x(3x-2)+4(3x-2)&\mbox{factor by grouping}\\
\displaystyle &\displaystyle= (5x+4)(3x-2)&\mbox{}\\
\end{array}
$$
$12x^2+8x u-15u^2$
$$
\begin{array}{c|c}
ac=12(-15)=-180 & \mbox{sum}\\ \hline
-1 \cdot 180 & 179\\
-2 \cdot 90 & 88\\
-3 \cdot 60 & 57 \\
-4 \cdot 30 & 26 \\
-5 \cdot 36 & 31 \\
-6 \cdot 30 & 24 \\
-9 \cdot 20 & 11 \\
-10 \cdot 18 & 8 \, \checkmark \\
\end{array}
$$
$$
\begin{array}{lll}
\displaystyle 12x^2+8x u-15u^2 &\displaystyle= 12x^2-10xu+18x u-15u^2&\mbox{}\\
\displaystyle &\displaystyle=2x(6x-5u)+3u(6x-5u) &\mbox{}\\
\displaystyle &\displaystyle=(2x+3u)(6x-5u) &\mbox{}\\
\end{array}
$$
$3q^2-25q+12$
$$
\begin{array}{c|c}
ac=3(12)=36 & \mbox{sum}\\ \hline
(-1)(-36) & -37\\
(-2)(-18) & -20\\
(-3)(-12) & -15\\
(-4)(-9) & -13\\
(-6)(-6) & -12\\
\end{array}
$$
If we were to keep going, we would simply start recycling factors.
So, we're out of possibilities.
When this happens, we can say that the polynomial is not factorable.
So, we're out of possibilities.
When this happens, we can say that the polynomial is not factorable.