Introduction
Most of lecturers keep struggling to teach solving quadratic equation by factoring? Do your students just not “get it”? Well, it’s time to quit. Yep! You can teach (and do) factoring without taking much effort and time to teach them.
How to - Solving quadratic equation : Solving by Factoring?
To solve a quadratic equation by factoring:
• Transform the equation into standard form.
• Factor the quadratic expression by putting two brackets.
• Set each factor containing the variable equal to 0.
• Solve each of the resulting equations.
This method will not work for some of quadratic equations (cannot be factored or have complex solution).

Example : Solve the following by factoring quadratic equation

$\begin{array}{l}a\right)\text{\hspace{0.17em}}{x}^{2}-12x+35=0\\ \left[\begin{array}{l}{x}^{2}-12x+35=0\\ \left(x-5\right)\left(x-7\right)=0\\ \text{}x-5=0,\text{\hspace{0.17em}}x-7=0\\ \text{}\underset{¯}{\underset{¯}{x=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=7}}\text{}\end{array}\right]to\text{\hspace{0.17em}}factor〉\left(\begin{array}{cc}\begin{array}{l}x\\ \underset{¯}{x}\\ \underset{x×x}{\underbrace{{x}^{2}}}\end{array}& \begin{array}{l}-5\\ \underset{¯}{-7}\\ \underset{-5×-7}{\underbrace{35}}\end{array}\end{array}|\begin{array}{c}-5x\left\{cross\text{\hspace{0.17em}}multiply\text{\hspace{0.17em}}x×-5\\ \underset{¯}{-7x\left\{cross\text{\hspace{0.17em}}multiply\text{\hspace{0.17em}}x×-7}\\ \underset{\left(-5x\right)+\left(-7x\right)}{\underbrace{-12x}}\end{array}\right)\end{array}$

$\begin{array}{l}b\right)\frac{x+5}{2}=\frac{12}{x}\\ \left[\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}x\left(x+5\right)\text{}=2×12\text{}\text{}\text{}\text{}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}+5x\text{}=24\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}+5x-24\text{}=0\\ \left(x+8\right)\left(x-3\right)\text{}=0\\ \text{}x+8=0\text{},\text{\hspace{0.17em}}x-3=0\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{¯}{\underset{¯}{x=-8\text{},\text{\hspace{0.17em}}x=3}}\text{}\end{array}\right]to\text{\hspace{0.17em}}factor〉\left(\begin{array}{cc}\begin{array}{l}x\\ \underset{¯}{x}\\ \underset{x×x}{\underbrace{{x}^{2}}}\end{array}& \begin{array}{l}8\\ \underset{¯}{-3}\\ \underset{8×-3}{\underbrace{-24}}\end{array}\end{array}|\begin{array}{c}8x\left\{cross\text{\hspace{0.17em}}multiply\text{\hspace{0.17em}}x×8\\ \underset{¯}{-3x\left\{cross\text{\hspace{0.17em}}multiply\text{\hspace{0.17em}}\text{\hspace{0.17em}}x×-3}\\ \underset{\left(8x\right)+\left(-3x\right)}{\underbrace{5x}}\end{array}\right)\end{array}$

$\begin{array}{l}c\right)\text{\hspace{0.17em}}2{x}^{2}+6x+4={x}^{2}-4x-12\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2{x}^{2}-{x}^{2}+6x+4x+4+12=0\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{\hspace{0.17em}}{x}^{2}+10x+16=0\\ \left(x+8\right)\left(x+2\right)=0\\ \text{\hspace{0.17em}}x+8=0,\text{\hspace{0.17em}}x+2=0\\ \text{\hspace{0.17em}}\underset{¯}{\underset{¯}{x=-8\text{},\text{\hspace{0.17em}}x=-2}}\end{array}\right]to\text{\hspace{0.17em}}factor〉\left(\begin{array}{cc}\begin{array}{l}x\\ \underset{¯}{x}\\ \underset{x×x}{\underbrace{{x}^{2}}}\end{array}& \begin{array}{l}8\\ \underset{¯}{2}\\ \underset{8×2}{\underbrace{16}}\end{array}\end{array}|\begin{array}{c}8x\left\{cross\text{\hspace{0.17em}}multiply\text{\hspace{0.17em}}x×8\\ \underset{¯}{2x\left\{cross\text{\hspace{0.17em}}multiply\text{\hspace{0.17em}}\text{\hspace{0.17em}}x×2}\\ \underset{\left(8x\right)+\left(2x\right)}{\underbrace{10x}}\end{array}\right)\end{array}$

Here are three examples: Bear in mind that positive and negative sign of the roots play important role in this method.

Try it yourself - Solving Quadratic Equation : Solving by Factoring
$\begin{array}{l}\begin{array}{l}a\right){x}^{2}-4x+3=0\\ b\right){x}^{2}+10x+9=0\\ c\right)\text{\hspace{0.17em}}2{x}^{2}-x-1=0\end{array}\right\}\begin{array}{c}ans-x=-1,-3\\ ans-x=1,9\\ ans-x=1,-\frac{1}{2}\end{array}\\ \end{array}$

More Exercise

Other methods

What do you think - Students?

Can you learn factoring the quadratic equation this way? Share your thoughts and experiences in the comment.