Section 4.1
Solving Quadratics by Factoring
The first way to solve a quadratic equation is by factoring. If a quadratic equation is in standard form, and if it factors, then its solutions are easy to determine. If the product of two numbers is 0, then one or the other of the numbers must be 0. We'll call this the Zero Product Rule.$$\text{If }A\cdot B=0\text{ then either } A=0 \text{ or } B=0.$$
Examples:
a. If $x^2-7x+12=0$, then $(x-3)(x-4)=0$ The only way that can occur is if either $x-3=0$ or $x-4=0$. Hence the answer is either $x=3$ or $x=4$.
b. Solve for $x$: $2x^2-3x+1=0$
Solution:
\begin{align}
2x^2-3x+1&=0\\
(2x-1)(x-1)&=0&&\textit{Factor the quadratic first.}\\
2x-1=0\;\;\text{o}&\text{r }\;\;x-1=0&&\textit{Separate into two equations using the Zero Product Rule.}\\
\text{So, }\; x=\frac{1}{2}\;\;\text{o}&\text{r}\;\;x=1&&\textit{Solve each equation separately.}
\end{align}
c. Solve for $x$: $x^2+x=20$
Solution:
\begin{align}
x^2+x&=20\\
x^2+x-&=20&&\textit{Put the quadratic in standard form.}\\
(x+5)(x-4)&=0&&\textit{Factor.}\\
x+5=0\;\;\text{o}&\text{r}\;\;x-4=0&&\textit{Separate into two equations using the Zero Product Rule.}\\
\text{So, }\;x=-5\;\;\text{o}&\text{r}\;\;x=4&&\textit{Solve each equation separately.}
\end{align}
Exercises
Solve each of the following for $x$.\begin{array}{l l l}
1. \;\;{{x}^{2}}-5x-6=0\;\;& 2. \;\;{{x}^{2}}+8x+12=0\;\;& 3. \;\;{{x}^{2}}-4x-21=0\;\; \\
4. \;\;{{x}^{2}}-8x-48=0\;\;& 5. \;\;{{x}^{2}}+3x-10=0\;\;& 6. \;\;{{x}^{2}}-4=0\;\; \\
7. \;\;{{x}^{2}}-8x=0\;\;& 8. \;\;{{x}^{2}}-3x-28=0\;\;& 9. \;\;{{x}^{2}}+4x+3=0\;\; \\
10. \;\;2{{x}^{2}}-x-1=0\;\;& 11. \;\;3{{x}^{2}}+5x+2=0\;\;& 12. \;\;3{{x}^{2}}+5x-2=0\;\;
\end{array}
\begin{array}{l l}
13. \;\;(x-1)(x+2)=(x-1)(5x+4)\;\;& 14. \;\;(x+4)(x-1)=(x+4)(2x-5)\;\; \\
15. \;\;(3x-1)(5x+2)=(3x-1)(x+3)\;\;& 16. \;\;(x+4)(2x-1)=(3x+4)(2x-1)\;\; \\
17. \;\;{{x}^{2}}-5x-6={{x}^{2}}-1+{{(x+1)}^{2}}\;\;& 18. \;\;{{x}^{2}}-x-12=(x+3)(x+4)+{{x}^{2}}-9\;\; \\
19. \;\;3{{x}^{2}}+2x-1=(x+1)(x+4)+{{(x+1)}^{2}}\;\;& 20. \;\;24{{x}^{2}}+x=10\;\; \\
21. \;\;2{{x}^{2}}+11x+15=0\;\;& 22. \;\;3{{x}^{2}}-4x+1=(x+1)(x-1)\;\; \\
23. \;\;{{x}^{2}}-3ax=4{{a}^{2}}\;\;& 24. \;\;(x+1)(x+2)+(2x+3)(3x+4)=14\;\; \\
25. \;\;x(x-a)=12{{a}^{2}}\;\;& 26. \;\;x(x+2a)=8{{a}^{2}}\;\;
\end{array}