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

by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!

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

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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