Section 2.6
Trinomials — $ax^2\pm bx \pm c$
We now consider more general quadratic trinomials — ones with a leading coefficient that is not zero.
Consider an expression such as $$6x^2-17x+12\;\;\;\;\;\text{ and }\;\;\;\;\;6x^2-x-12.$$ Neither of these is anywhere near as easy to factor by inspection as $x^2-x-12$, because the leading coefficient, 6, isn't prime, and so the number of combinations to consider is vast. What follows is an explanation of a factoring technique to consider when confronted with these problems.
a. To factor $6x^2-17x+12$, we take the product of 6 and 12, the leading coefficient and the constant term. This product is 72. Since the 12 has a plus sign, we look for two numbers whose product is 72 and whose sum is 17 (the middle coefficient, ignoring the sign). The two numbers that work are 8 and 9. Now we adjust the signs; since the 17 has a minus sign, we actually need $-8$ and $-9$.
Rewrite the quadratic by replacing the middle term: $$6x^2-17x+12=6x^2-8x-9x+12.$$
Now we use the method of grouping "two and two" to factor this four-term expression
\begin{align}
6x^2-17x+12&=6x^2-8x-9x+12\\
&=(6x^2-8x)-(9x-12)\\
&=2x(3x-4)-3(3x-4)\\
&=(2x-3)(3x-4)
\end{align}
There is a second way we could have replaced the middle term. We could have written the two middle terims in the other order:
\begin{align}
6x^2-17x+12&=6x^2-9x-8x+12\\
&=(6x^2-9x)-(8x-12)\\
&=3x(2x-3)-4(2x-3)\\
&=(2x-3)(3x-4)
\end{align}
b. To factor $6x^2-x-12$, we start by considering the product of the 6 and the 12 (as earlier, we ignore signs for the moment). This time, since the 12 has a minus sign, we look for two numbers whose product is 72 and whose difference is 1 (the linear coefficient, ignoring signs). Again the numbers are 8 and 9. Since we need $-1x$ in the middle, we use $+8x$ and $-9x$. Recommended is to use the "minus" term first. Replace the middle term with these two new terms, and group "two and two."
\begin{align}
6x^2-x-12&=6x^2-9x+8x-12\\
&=(6x^2-9x)+(8x-12)\\
&=3x(2x-3)+4(2x-3)\\
&=(3x+4)(2x-3)
\end{align}
This will work if we write the middle terms in the other order, but there are increased dangers of making an error with the signs.
Factor by grouping and factoring completely: Factor by Grouping and Factoring Completely
Example 1: Factoring by grouping: Factoring trinomials with a non-1 leading coefficient by grouping
Exercises:
Factor each of the following.
\begin{array}{ l l l }
1. \;\;6{{x}^{2}}+23x+20\;\;& 2. \;\;15{{x}^{2}}-23x+6\;\;& 3. \;\;6{{x}^{2}}-19x+8\;\;\\
4. \;\;6{{x}^{2}}-x-15\;\;& 5. \;\;6{{x}^{2}}+3x-9\;\;& 6. \;\;9{{x}^{2}}+3x-20\;\;\\
7. \;\;12{{x}^{2}}-17x+6\;\;& 8. \;\;8{{x}^{2}}+26x+15\;\;& 9. \;\;8{{x}^{2}}+14x-15\;\;\\
10. \;\;6{{x}^{2}}+7x-20\;\;& 11. \;\;6{{x}^{2}}+6x-12\;\;& 12. \;\;18{{x}^{2}}-23x-6\;\;\\
13. \;\;8{{x}^{2}}+10x-7\;\;& 14. \;\;12{{x}^{2}}+x-6\;\;& 15. \;\;20{{x}^{2}}+7x-6\;\;\\
16. \;\;8{{x}^{2}}+30x+7\;\;& 17. \;\;15{{x}^{2}}-22x+8\;\;& 18. \;\;20{{x}^{2}}-x-12\;\;\\
19. \;\;15{{x}^{2}}-2x-8\;\;& 20. \;\;18{{x}^{2}}+51x+8\;\;& 21. \;\;12{{x}^{2}}+37x+21\;\;
\end{array}