Using Factoring to Solve Quadratic Equations
| Site: | Saylor University |
| Course: | [CB DRAFT] MA007: Real World Math: Algebra |
| Book: | Using Factoring to Solve Quadratic Equations |
| Printed by: | Guest user |
| Date: | Tuesday, 27 October 2026, 4:20 PM |
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 $$x=\frac{4}{3}$$