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Solving Quadratics by Factoring,
Factoring Quadratic Equations


As seen on the quadratic equations introduction page, quadratic equations have the following general form.

ax^{\tt{2}} + bx + c = 0

The factor by grouping page showed how to effectively factor quadratics when possible, this page will show how this approach can be used to solve quadratics by factoring.




Solutions of a Quadratic Equation


Let’s consider a basic example of a quadratic equation.    x^{\tt{2}} + x \space {\text{–}} \space 2 = 0

This quadratic equation factors nicely.    (x + 2)(x \space {\text{–}} \space 1) = 0


If we set both binomials equal to zero and solve, we obtain two  x  values.

x + 2 = 0     =>     x = {\text{-}}2
x \space {\text{–}} \space 1 = 0      =>     x = 1

Now if we look at the graph of   y =x^{\tt{2}} + x \space {\text{–}} \space 2,
we can see what these  x  values represent.

Graph helping show how to solve quadratics by factoring.


The  x  values found are the solutions of the quadratic equation. Often also called ‘roots’ or ‘zeroes.
The values with which the graph touches the  x-axis.


Usually it’s good practice to input any root values found into the equation to check that they do definitely hold. When solving quadratics by factoring.

({\text{-}}2)^2 + ({\text{-}}2) \space {\text{–}} \space 2 \space = \space 4 \space {\text{–}} \space 2 \space {\text{–}} \space 2 \space = \space 0    ( CORRECT )

(1)^2 + 1 \space {\text{–}} \space 2 \space = 1 + 1 \space {\text{–}} \space 2 \space = \space 0    ( CORRECT )




Solving Quadratics by Factoring Summary:

–   With the quadratic equation equal to zero, factor the quadratic expression on the left hand side.

–   Set the factor terms found equal to zero, and solve them.

–   Double check the solutions obtained by inputting into the original quadratic equation.






Solving Quadratics by Factoring Examples



(1.1) 

Solve   x^{\tt{2}} + 7x \space {\text{–}} \space 8 = 0.

Solution   

x^{\tt{2}} + 7x \space {\text{–}} \space 8 = 0
factors to    (x \space {\text{–}} \space 1)(x + 8) = 0.

(x \space {\text{–}} \space 1) = 0    =>    x = 1
(x + 8) = 0    =>    x = {\text{-}}8

Now to check the found solutions.

x = 1 :     1^{\tt{2}} + 7(1) \space {\text{–}} \space 8   =   1 + 7 \space {\text{–}} \space 8 = 0

x = {\text{-}}8 :     ({\text{-}}8)^{\tt{2}} + 7({\text{-}}8) \space {\text{–}} \space 8   =   64 \space {\text{–}} \space 56 \space {\text{–}} \space 8 = 0




(1.2) 

Solve   4x^{\tt{2}} + x \space {\text{–}} \space 5 = 0.

Solution   

4x^{\tt{2}} + x \space {\text{–}} \space 5 = 0
factors to    (4x + 5)(x \space {\text{–}} \space 1) = 0.

(4x + 5) = 0    =>    x = {\text{-}}{\tt{\frac{5}{4}}}
(x \space {\text{–}} \space 1) = 0    =>    x = 1

Now to check the found solutions.

x = {\text{-}}{\tt{\frac{5}{4}}} :     4({\text{-}}{\tt{\frac{5}{4}}})^{\tt{2}} + ({\text{-}}{\tt{\frac{5}{4}}}) \space {\text{–}} \space 5   =   4({\tt{\frac{25}{16}}}) \space {\text{–}} \space {\tt{\frac{5}{4}}} \space {\text{–}} \space 5 = 0   =   {\tt{\frac{100}{16}}} \space {\text{–}} \space {\tt{\frac{20}{16}}} \space {\text{–}} \space {\tt{\frac{80}{16}}} = 0

x = 1 :     4(1)^{\tt{2}} + 1 \space 5   =   4 + 1 \space {\text{–}} \space 5 = 0




(1.3) 

Solve   4x^{\tt{2}} + 12x + 9 = 0.

Solution   

4x^{\tt{2}} + 12x + 9 = 0
factors to    (2x + 3)(2x + 3) = 0.

(2x + 3) = 0    =>    x = {\text{-}}{\tt{\frac{3}{2}}}

Two identical roots means there is just one solution overall, which can be called a ‘double root’.

x = {\text{-}}{\tt{\frac{3}{2}}} :     4({\text{-}}{\tt{\frac{3}{2}}})^{\tt{2}} + 12({\text{-}}{\tt{\frac{3}{2}}}) + 9    =    4({\tt{\frac{9}{4}}}) \space {\text{–}} \space {\tt{\frac{36}{2}}} + 9    =    {\tt{\frac{36}{4}}} \space {\text{–}} \space {\tt{\frac{72}{4}}} + {\tt{\frac{36}{4}}} = 0





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