Using Factoring to Solve Complicated 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

Solve: $3x^{2}=5-14x$

Solution: Write a nice, clean list of equivalent 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.

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.

Check by substituting into the original equation:

$10(-\frac{2}{3})^{2}-11(-\frac{2}{5})-6=0$;  $10(\frac{4}{25})+\frac{22}{5}-6=0$;  $2(\frac{4}{5})+\frac{22}{5}-\frac{30}{5}=0$;  $0=0$;  Check!

$10(\frac{3}{2})^{2}-11(\frac{3}{2})-6=0$;  $10(\frac{9}{4})-\frac{33}{2}-6=0$;  $5(\frac{9}{2})-\frac{33}{2}-\frac{12}{2}=0$; $0=0$;  Check!

Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/solve_quad_eq_morecomp_fac.htm
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