﻿ Solving More Complicated Quadratic Equations by Factoring
SOLVING MORE COMPLICATED QUADRATIC EQUATIONS BY FACTORING

To solve a quadratic equation by factoring:

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!
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Multiplying and Dividing Fractions with Variables

CONCEPT QUESTIONS EXERCISE:

For more advanced students, a graph is displayed.
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 if you prefer not to see the graph.

PROBLEM TYPES:
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