More on Using Factoring to Solve Quadratic Equations
| Site: | Saylor Academy |
| Course: | MA007: Real World Math: Algebra |
| Book: | More on Using Factoring to Solve Quadratic Equations |
| Printed by: | Guest user |
| Date: | Friday, November 28, 2025, 6:08 AM |
Description
buttons.More on Using Factoring to Solve Quadratic Equations
To solve a quadratic equation by factoring:
- Put it in standard form: \(ax^{2}+bx+c=0\)
- Factor the left-hand side
- Use the Zero Factor Law
Examples
Solve: \(3x^{2}=5-14x\)
Solution:
Write a nice, clean list of equivalent equations.
| \(3x^{2}=5-14x\) | Original equation |
| \(3x^{2}+14x-5=0\) | Put in standard form: subtract \(5\) from both sides; add \(14x\) to both sides |
| \((3x-1)(x+5)=0\) | Factor the left-hand side; you may want to use the factor by grouping method |
| \(3x-1=0\) or \(x+5=0\) | Use the Zero Factor Law |
| \(3x=1\) or \(x=-5\) | Solve the simpler equations |
| \(x=\frac{1}{3}\) or \(x=-5\) | Solve the simpler equations |
Check by substituting into the original equation:
\(3(\frac{1}{3})^{2}\stackrel {?} {=}5-14(\frac{1}{3})\)
\(3\cdot \frac{1}{9}\stackrel {?} {=}\frac{15}{3}-\frac{14}{3}\)
\(\frac{1}{3}=\frac{1}{3}\)
Check!
\(3(-5)^{2}\stackrel {?} {=}5-14(-5)\)
\(3\cdot 25\stackrel {?} {=}5+70\)
\(75=75\)
Check!
Solve: \((2x+3)(5x-1)=0\)
Solution:
Do not multiply it out!
If it is already in factored form, with zero on one side, then be happy that a lot of the work has already been done for you.
| \((2x+3)(5x-1)=0\) | Original equation |
| \(2x+3=0\) or \(5x-1=0\) | Use the Zero Factor Law |
| \(2x=-3\) or \(5x=1\) | Solve the simpler equations |
| \(x=-\frac{3}{2}\) or \(x=\frac{1}{5}\) | Solve the simpler equations |
Check by substituting into the original equation:
\((2(-\frac{3}{2})+3)(5(-\frac{3}{2})-1)\stackrel {?} {=}0\)
\(0=0\)
Check!
\((2(\frac{1}{5})+3)(5(\frac{1}{5})-1)\stackrel {?} {=}0\)
\(0=0\)
Check!
Solve: \(10x^{2}-11x-6=0\)
Solution:
Note that it is already in standard form.
| \(10x^{2}-11x-6=0\) | Original equation |
| \((5x+2)(2x-3)=0\) | Factor the left-hand side; you may want to use the factor by grouping method |
| \(5x+2=0\) or \(2x-3=0\) | Use the Zero Factor Law |
| \(5x=-2\) or \(2x=3\) | Solve the simpler equations |
| \(x=-\frac{2}{5}\) or \(x=\frac{3}{2}\) | Solve the simpler equations |
Check by substituting into the original equation:
\(10(-\frac{2}{3})^{2}-11(-\frac{2}{5})-6\stackrel {?} {=}0\)
\(10(\frac{4}{25})+\frac{22}{5}-6\stackrel {?} {=}0\)
\(2(\frac{4}{5})+\frac{22}{5}-\frac{30}{5}\stackrel {?} {=}0\)
\(0=0\)
Check!
\(10(\frac{3}{2})^{2}-11(\frac{3}{2})-6\stackrel {?} {=}0\)
\(10(\frac{9}{4})-\frac{33}{2}-6\stackrel {?} {=}0\)
\(5(\frac{9}{2})-\frac{33}{2}-\frac{12}{2}\stackrel {?} {=}0\)
\(0=0\)
Check!
Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/solve_quad_eq_morecomp_fac.htm
This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License.
Practice Problems
Solve. Write a nice, clean list of equivalent equations.
\(0=4x^2+2x−2\)
\(4=−7x+2x^2\)
\(15x=−9x^2−6\)
\((4x)(−3x−5)=0\)
Answers
\(x=−1\) or \(x=\frac{1}{2}\)
\(x=4\) or \(x=−\frac{1}{2}\)
\(x=−\frac{2}{3}\) or \(x=−1\)
\(x=0\) or \(x=−\frac{5}{3}\)