# Factoring method of solving Quadratic Equations

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.

## Method

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.

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

### Proof

$ax^2+bx+c = 0$ 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.

#### Example

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.

##### Step: 1

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.

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

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

##### Step: 2

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

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

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

##### Step: 3

$x+3 = 0$ 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.

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