Using Factoring to Solve Quadratic Equations
| Site: | Saylor Academy |
| Course: | MA007: Algebra |
| Book: | Using Factoring to Solve Quadratic Equations |
| Printed by: | Guest user |
| Date: | Tuesday, 15 July 2025, 7:37 AM |
Description
buttons.Using Factoring to Solve Quadratic Equations
To solve a quadratic equation by factoring:
- Put it in standard form: \(ax2+bx+c=0\)
- Factor the left-hand side
- Use the Zero Factor Law
Examples
Solve: \(x^{2}=2-x\)
Solution:
Write a nice, clean list of equivalent equations.
| \(x^{2}=2-x\) | Original equation |
| \(x^{2}+x-2=0\) | Put in standard form: subtract \(2\) from both sides; add \(x\) to both sides |
| \((x+2)(x-1)=0\) | Factor the left-hand side |
| \(x+2=0\) or \(x-1=0\) | Use the Zero Factor Law |
| \(x=-2\) or \(x=1\) | Solve the simpler equations |
Check by substituting into the original equation:
\((-2)^{2} \stackrel {?} {=}2-(-2)\)
\(4=4\);
Check!
\((1)^{2}\stackrel {?} {=}2-1\)
\(1=1\)
Check!
Solve: \((x+3)(x-2)=0\)
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.
| \((x+3)(x-2)=0\) | Original equation |
| \(x+3=0\) or \(x-2=0\) | Use the Zero Factor Law |
| \(x=-3\) or \(x=2\) | Solve the simpler equations |
Check by substituting into the original equation:
\((-3+3)(-3-2)\stackrel {?} {=}0\)
\(0=0\)
Check!
\((2+3)(2-2)\stackrel {?} {=}0\)
\(0=0\)
Check!
Solve: \((2x-3)(1-3x)=0\)
Solution:
Again, do not multiply it out! When you have a product on one side, and zero on the other side, then you are all set to use the Zero Factor Law.
| \((2x-3)(1-3x)=0\) | Original equation |
| \(2x-3=0\) or \(1-3x=0\) | Use the Zero Factor Law |
| \(2x=3\) or \(1-3x\) | Solve simpler equations |
| \(x=\frac{3}{2}\) or \(x=\frac{1}{3}\) | Solve simpler equations |
Check by substituting into the original equation:
\((2\cdot \frac{3}{2}-3)(1-3\cdot \frac{3}{2})\stackrel {?} {=}0\)
\(0=0\)
Check!
\((2\cdot \frac{1}{3}+3)(1-3\cdot \frac{1}{3})\stackrel {?} {=}0\)
\(0=0\)
Check!
Solve: \(x^{2}+4x-5=0\)
Solution:
Note that it is already in standard form.
| \(x^{2}+4x-5=0\) | Original equation |
| \((x+5)(x-1)=0\) | Factor the left-hand side |
| \(x+5=0\) or \(x-1=0\) | Use the Zero Factor Law |
| \(x=-5\) or \(x=1\) | Solve the simpler equations |
Check by substituting into the original equation:
\((-5)^{2}+4(-5)-5\stackrel {?} {=}0\)
\(25-20-5\stackrel {?} {=}0\)
\(0=0\)
Check!
\(1^{2}+4(1)-5\stackrel {?} {=}0\)
\(1+4-5\stackrel {?} {=}0\)
\(0=0\)
Check!
Solve: \(14=-5x+x^{2}\)
Solution:
| \(14=-5x+x^{2}\) | Original equation |
| \(x^{2}-5x-14=0\) | Put in standard form: subtract \(14\) from both sides; write in the conventional way |
| \((x-7)(x+2)=0\) | Factor the left-hand side |
| \(x-7=0\) or \(x+2=0\) | Use the Zero Factor Law |
| \(x=7\) or \(x=-2\) | Solve the simpler equations |
Check by substituting into the original equation:
\(14\stackrel {?} {=}-5(7)+7^{2}\)
\(14\stackrel {?} {=}-35+49\)
\(14=14\)
Check!
\(14\stackrel {?} {=}-5(-2)+(-2)^{2}\)
\(14\stackrel {?} {=}10+4\)
\(14=14\)
Check!
Solve: \(6x=2x^{2}\)
Solution:
When there is no constant term, the factoring is much easier.
| \(6x=2x^{2}\) | Original equation |
| \(2x^{2}-6x=0\) | Put in standard form: subtract \(6x\) from both sides; write in the conventional way |
| \(x^{2}-3x=0\) | Optional step: divide both sides by \(2\) |
| \(x(x-3)=0\) | Factor the left-hand side |
| \(x=0\) or \(x-3=0\) | Use the Zero Factor Law |
| \(x=0\) or \(x=3\) | Solve the simpler equations |
Check by substituting into the original equation:
\(6\cdot 0\stackrel {?} {=}2\cdot 0^{2}\)
\(0=0\)
Check!
\(6\cdot 3\stackrel {?} {=}2\cdot 3^{2}\)
\(18=18\)
Check!
Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/solve_quad_eq_simple_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.
\(x^2=2x+8\)
\(15−8x=−x^2\)
\(−4x+x^2=−3\)
\(−20+x=−x^2\)
\(x^2+x−12=0\)
\(x^2=3x−2\)
\(x^2=−3x+10\)
\(x^2=−4x−3\)
\(−3−2x=−x2\)
\(5x+x^2=−4\)
\(x^2−3x+2=0\)
\(x^2−2x−15=0\)
\(x^2+10x+25=0\)
\(x+x^2=12\)
\(10+7x=−x^2\)
\(−x+x^2=12\)
\((x+1)(x−8)=0\)
\((10x−3)(3x−4)=0\)
Answers
\(x=4\) or \(x=−2\)
\(x=5\) or \(x=3\)
\(x=3\) or \(x=1\)
\(x=−5\) or \(x=4\)
\(x=−4\) or \(x=3\)
\(x=1\) or \(x=2\)
\(x=−5\) or \(x=2\)
\(x=−3\) or \(x=−1\)
\(x=3\) or \(x=−1\)
\(x=−1\) or \(x=−4\)
\(x=1\) or \(x=2\)
\( x=5\) or \(x=−3\)
\(x=−5\)
\(x=−4\) or \(x=3\)
\(x=−5\) or \(x=−2\)
\(x=4\) or \(x=−3\)
\(x=−1\) or \(x=8\)
\(x=\frac{3}{10}\) or \(x=\frac{4}{3}\)