Solving More Complicated Quadratic Equations by Factoring

Need some basic practice with quadratic equations first?

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
$3x-1 = 0\ \ \text{ or }\ \ x + 5 = 0$	use the Zero Factor Law
$3x = 1\ \ \text{ or }\ \ x = -5$	solve the simpler equations
$\displaystyle x = \frac{1}{3}\ \ \text{ or }\ \ x = -5$	solve the simpler equations

Check by substituting into the original equation:

$3{(\frac{1}{3})}^2 \ \ \overset{\text{?}}{=}\ \ 5 - 14(\frac13)$
$3\cdot\frac19 \overset{\text{?}}{=} \frac{15}{3}-\frac{14}3$
$\frac13 = \frac 13$
Check!

$3{(-5)}^2 \ \ \overset{\text{?}}{=}\ \ 5 - 14(-5)$
$3\cdot 25 \ \overset{\text{?}}{=}\ 5 + 70$
$75 = 75$
Check!

Solve: $(2x+3)(5x-1) = 0$

Solution: Note: Don't multiply it out! If it's 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
$2x+3 = 0\ \ \text{ or }\ \ 5x - 1 = 0$	use the Zero Factor Law
$2x = -3\ \ \text{ or }\ \ 5x = 1$	solve the simpler equations
$\displaystyle x = -\frac{3}{2}\ \ \text{ or }\ \ x = \frac{1}{5}$	solve the simpler equations

Check by substituting into the original equation:

$(2(-\frac32)+3)(5(-\frac32)-1) \ \ \overset{\text{?}}{=}\ \ 0$
$0 = 0$
Check!

$(2(\frac15)+3)(5(\frac15)-1) \ \ \overset{\text{?}}{=}\ \ 0$
$0 = 0$
Check!

Solve: $10x^2 - 11x - 6 = 0$

Solution: Note that it's 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
$5x+2 = 0\ \ \text{ or }\ \ 2x - 3 = 0$	use the Zero Factor Law
$5x = -2\ \ \text{ or }\ \ 2x = 3$	solve the simpler equations
$\displaystyle x = -\frac{2}{5}\ \ \text{ or }\ \ x = \frac{3}{2}$	solve the simpler equations

Check by substituting into the original equation:

$10(-\frac25)^2 - 11(-\frac25) - 6 \ \ \overset{\text{?}}{=}\ \ 0$
$10(\frac{4}{25}) + \frac{22}{5} - 6 \ \ \overset{\text{?}}{=}\ \ 0$
$2(\frac{4}{5}) + \frac{22}{5} - \frac{30}{5} \ \ \overset{\text{?}}{=}\ \ 0$
$0 = 0$
Check!

$10{(\frac32)}^2 - 11(\frac32) - 6 \ \ \overset{\text{?}}{=}\ \ 0$
$10(\frac{9}{4}) - \frac{33}{2} - 6 \ \ \overset{\text{?}}{=}\ \ 0$
$5(\frac{9}{2}) - \frac{33}{2} - \frac{12}{2} \ \ \overset{\text{?}}{=}\ \ 0$
$0 = 0$
Check!

Concept Practice

For more advanced students, a graph is available. For example, the equation $\,3x^2 = 5 - 14x\,$ is optionally accompanied by the graph of $\,y = 3x^2\,$ (the left side of the equation, dashed green) and the graph of $\,y = 5 - 14x\,$ (the right side of the equation, solid purple). In this example, you are finding the values of $\,x\,$ where the green graph intersects the purple graph.

Click the ‘Show/Hide Graph’ button to toggle the graph.