Using Factoring to Solve Complicated Quadratic Equations

Read each example slowly and try to identify the factoring methods being used and why each step is performed. After you have reviewed the materials, complete a few practice problems and check your answers.

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

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:

$3(\frac{1}{3})^{2}=5-14(\frac{1}{3})$ ; $3\cdot \frac{1}{9}=\frac{15}{3}-\frac{14}{3}$ ; $\frac{1}{3}=\frac{1}{3}$ ; Check!

$3(-5)^{2}=5-14(-5)$ ; $3\cdot 25=5+70$ ; $75=75$ 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:

$(2(-\frac{3}{2})+3)(5(-\frac{3}{2})-1)=0$ ; $0=0$ ; Check!

$(2(\frac{1}{5})+3)(5(\frac{1}{5})-1)=0$ ; $0=0$ ; 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