A method of solving a quadratic equation by writing the quadratic equation as two multiplying factors is called the factoring method of solving quadratic equations.

A quadratic equation is usually written as $ax^2+bx+c = 0$ in algebraic form. If this equation is transformed as two multiplying factors, then it is used to find the roots easily.

For example, if alpha ($\alpha$) and beta ($\beta$) are two roots of the quadratic equation, then the quadratic equation is written as follows.

$ax^2+bx+c = 0$ $\,\, \Leftrightarrow \,\,$ $a(x-\alpha)(x-\beta) = 0$

$ax^2+bx+c = 0$ is a quadratic equation in general form.

$\implies a\Bigg[x^2+\dfrac{bx}{a}+\dfrac{c}{a}\Bigg] = 0$

$\implies a\Bigg[x^2+\Bigg(\dfrac{b}{a}\Bigg)x+\dfrac{c}{a}\Bigg] = 0$

$\implies a\Bigg[x^2-\Bigg(\dfrac{-b}{a}\Bigg)x+\dfrac{c}{a}\Bigg] = 0$

The literal factor $\dfrac{-b}{a}$ represents the sum of the roots and the term $\dfrac{c}{a}$ represents the product of the roots of the quadratic equation $ax^2+bx+c = 0$. So, express the literal coefficient of $x$ as the sum of the roots and also express the third term of the quadratic equation as the product of the roots.

$\implies a\Bigg[x^2-\Bigg(\dfrac{-2b}{2a}\Bigg)x+\dfrac{c}{a}\Bigg] = 0$

$\implies a\Bigg[x^2-\Bigg(\dfrac{-b}{2a}+\dfrac{-b}{2a}\Bigg)x+\dfrac{c}{a}\Bigg] = 0$

$\implies a\Bigg[x^2-\Bigg(\dfrac{-b}{2a}$ $+$ $\dfrac{-b}{2a} +\dfrac{\sqrt{b^2 -4ac}}{2a} -\dfrac{\sqrt{b^2 -4ac}}{2a} \Bigg)x$ $+$ $\dfrac{c}{a}\Bigg] = 0$

$\implies a\Bigg[x^2-\Bigg(\dfrac{-b}{2a}$ $+$ $\dfrac{\sqrt{b^2 -4ac}}{2a} +\dfrac{-b}{2a} -\dfrac{\sqrt{b^2 -4ac}}{2a} \Bigg)x$ $+$ $\dfrac{c}{a}\Bigg] = 0$

Multiply both $\dfrac{-b}{2a}+\dfrac{\sqrt{b^2 -4ac}}{2a}$ and $\dfrac{-b}{2a} -\dfrac{\sqrt{b^2 -4ac}}{2a}$ and observe their product.

$\Bigg(\dfrac{-b}{2a} + \dfrac{\sqrt{b^2 -4ac}}{2a}\Bigg)$ $\times$ $\Bigg(\dfrac{-b}{2a} -\dfrac{\sqrt{b^2 -4ac}}{2a}\Bigg)$ $=$ ${\Bigg(\dfrac{-b}{2a}\Bigg)}^2$ $-$ ${\Bigg(\dfrac{\sqrt{b^2 -4ac}}{2a}\Bigg)}^2$

$=$ $\dfrac{b^2}{4a^2}$ $-$ $\dfrac{b^2 -4ac}{4a^2}$

$=$ $\dfrac{b^2 -b^2 +4ac}{4a^2}$

$=$ $\dfrac{4ac}{4a^2}$

$=$ $\dfrac{c}{a}$

$\implies \dfrac{c}{a}$ $=$ $\Bigg(\dfrac{-b}{2a} + \dfrac{\sqrt{b^2 -4ac}}{2a}\Bigg)$ $\times$ $\Bigg(\dfrac{-b}{2a} -\dfrac{\sqrt{b^2 -4ac}}{2a}\Bigg)$

Hence, the term $\dfrac{c}{a}$ can be expressed as the product of them in simplifying quadratic equation.

$\implies a\Bigg[x^2 -\Bigg( \dfrac{-b+\sqrt{b^2 -4ac}}{2a} $+$ \dfrac{-b-\sqrt{b^2 -4ac}}{2a} \Bigg)x$ $+$ $\Bigg(\dfrac{-b+\sqrt{b^2 -4ac}}{2a}\Bigg)$$\Bigg(\dfrac{-b-\sqrt{b^2 -4ac}}{2a}\Bigg)\Bigg] = 0$

Take $\alpha = \dfrac{-b+\sqrt{b^2 -4ac}}{2a}$ and $\beta = \dfrac{-b-\sqrt{b^2 -4ac}}{2a}$.

$\implies a[x^2 – (\alpha+\beta)x + \alpha \beta] = 0$

$\implies a[x^2 – \alpha x -\beta x + \alpha \beta] = 0$

$\implies a[x(x-\alpha) -\beta(x-\alpha)] = 0$

$\therefore \,\,\,\,\,\, a(x-\alpha)(x-\beta) = 0$

The equation is in the form of multiplying factors. Hence, this method of solving quadratic equation is called factoring method. Therefore, $a \ne 0$ but $x-\alpha = 0$ and $x-\beta = 0$ are solutions of the quadratic equation by factoring method.

A quadratic equation can be solved in factoring method in three simple steps.

$x^2+8x+15 = 0$ is an example quadratic equation and observe the steps of solving it by factoring method.

Express the numerical coefficient of second term as the sum of two numbers and also express the third term as the product of the same two numbers.

$\implies x^2+(3+5)x+3 \times 5 = 0$

$\implies x^2+3x+5x+3 \times 5 = 0$

Write the quadratic equation as the product of two factors by taking common factors out.

$\implies x(x+3) + 5(x+3) = 0$

$\implies (x+3)(x+5) = 0$

$x+3 = 0$ and $x+5 = 0$. Therefore, $x = -3$ and $x = -5$ are solutions of this quadratic equation by factoring method and the roots are $x = \{-3, -5\}$.

Practice problems in our worksheet of solving quadratic equations by factoring method and get complete knowledge on this topic.