To solve a quadratic equation by factoring:

  1. Put it in standard form: \(ax2+bx+c=0\)

  2. Factor the left-hand side

  3. Use the Zero Factor Law


Solve: \(x^{2}=2-x\) 

Solution: Write a nice, clean list of equivalent equations.

\(x^{2}=2-x\) Original equation
\(x^{2}+x-2=0\) Put in standard form: subtract \(2\) from both sides; add \(x\) to both sides
\((x+2)(x-1)=0\) Factor the left-hand side
\(x+2=0\) or \(x-1=0\) Use the Zero Factor Law
\(x=-2\) or \(x=1\) Solve the simpler equations


Check by substituting into the original equation:

\((-2)^{2}=2-(-2)\); \(4=4\); Check!

\((1)^{2}=2-1\); \(1=1\) Check!


Solve: \((x+3)(x-2)=0\)

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.

\((x+3)(x-2)=0\) Original equation
\(x+3=0\) or \(x-2=0\) Use the Zero Factor Law
\(x=-3\) or \(x=2\) Solve the simpler equations


Check by substituting into the original equation:

\((-3+3)(-3-2)=0\)\(0=0\) Check!

\((2+3)(2-2)=0\); \(0=0\) Check!


Solve: \((2x-3)(1-3x)=0\)

Solution: Again, do not multiply it out! When you have a product on one side, and zero on the other side, then you are all set to use the Zero Factor Law.

\((2x-3)(1-3x)=0\) Original equation
\(2x-3=0\) or \(1-3x=0\) Use the Zero Factor Law
\(2x=3\) or \(1-3x\) Solve simpler equations
\(x=\frac{3}{2}\) or \(x=\frac{1}{3}\) Solve simpler equations



Check by substituting into the original equation:

\((2\cdot \frac{3}{2}-3)(1-3\cdot \frac{3}{2})=0\); \(0=0\) Check!

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


Solve: \(x^{2}+4x-5=0\)

Solution: Note that it is already in standard form.

\(x^{2}+4x-5=0\) Original equation
\((x+5)(x-1)=0\) Factor the left-hand side
\(x+5=0\) or \(x-1=0\) Use the Zero Factor Law
\(x=-5\) or \(x=1\) Solve the simpler equations


Check by substituting into the original equation:

\((-5)^{2}+4(-5)-5=0\); \(25-20-5=0\); \(0=0\) Check!

\(1^{2}+4(1)-5+0\); \(1+4-5=0\); \(0=0\) Check!


Solve: \(14=-5x+x^{2}\)

Solution:

\(14=-5x+x^{2}\) Original equation
\(x^{2}-5x-14=0\) Put in standard form: subtract \(14\) from both sides; write in the conventional way
\((x-7)(x+2)=0\) Factor the left-hand side
\(x-7=0\) or \(x+2=0\) Use the Zero Factor Law
\(x=7\) or \(x=-2\) Solve the simpler equations


Check by substituting into the original equation:

\(14=-5(7)+7^{2}\); \(14=-35+49\); \(14=14\) Check!

\(14=-5(-2)+(-2)^{2}\); \(14=10+4\); \(14=14\) Check!


Solve: \(6x=2x^{2}\)

Solution: When there is no constant term, the factoring is much easier.

\(6x=2x^{2}\) Original equation
\(2x^{2}-6x=0\) Put in standard form: subtract \(6x\) from both sides; write in the conventional way
\(x^{2}-3x=0\) Optional step: divide both sides by \(2\)
\(x(x-3)=0\) Factor the left-hand side
\(x=0\) or \(x-3=0\) Use the Zero Factor Law
\(x=0\) or \(x=3\) Solve the simpler equations


Check by substituting into the original equation:

\(6\cdot 0=2\cdot 0^{2}\); \(0=0\); Check!

\(6\cdot 3=2\cdot 3^{2}\); \(18=18\); Check!



Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/solve_quad_eq_simple_fac.htm
Creative Commons License This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License.

Last modified: Wednesday, May 5, 2021, 2:31 PM