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More on Using Factoring to Solve Quadratic Equations

Site: Saylor University
Course: [CB DRAFT] MA007: Real World Math: Algebra
Book: More on Using Factoring to Solve Quadratic Equations
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Date: Tuesday, 27 October 2026, 2:40 PM

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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
Creative Commons License This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License.

Practice Problems

Solve. Write a nice, clean list of equivalent equations.

  1. $$0=4x^2+2x−2$$

  2. $$4=−7x+2x^2$$

  3. $$15x=−9x^2−6$$

  4. $$(4x)(−3x−5)=0$$

Answers

  1. $$x=−1$$ or $$x=\frac{1}{2}$$

  2. $$x=4$$ or $$x=−\frac{1}{2}$$

  3. $$x=−\frac{2}{3}$$ or $$x=−1$$

  4. $$x=0$$ or $$x=−\frac{5}{3}$$