A mathematical approach of converting the quadratic expression into a product of two factors is called the factorization (or factorisation) of the quadratic equation.

A quadratic expression is a second degree polynomial. So, it has two roots. The idea of converting a quadratic expression as a product of two linear expressions in one variable helps us to find the zeros (or roots) of any quadratic equation easily. This method is called the factorization (or factorisation) of quadratic equations.

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

Now, take the literal coefficient of $x^2$ common from all the terms in the quadratic equation.

$\implies$ $a\bigg(x^2+\Big(\dfrac{b}{a}\Big)x+\dfrac{c}{a}\bigg) \,=\, 0$

Completing the square method is used to derive a quadratic formula for finding the roots of a quadratic equation.

$x \,=\, \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\implies$ $x \,=\, \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ or $x \,=\, \dfrac{-b-\sqrt{b^2-4ac}}{2a}$

If the roots of a quadratic equation are denoted by alpha and beta, then its solutions are expressed as follows.

$(1).\,\,\,$ $\alpha \,=\, \dfrac{-b+\sqrt{b^2-4ac}}{2a}$

$(2).\,\,\,$ $\beta \,=\, \dfrac{-b-\sqrt{b^2-4ac}}{2a}$

The sum and product of the roots of the quadratic equations are expressed as follows.

$(1).\,\,\,$ $\alpha+\beta \,=\, -\dfrac{b}{a}$

$(2).\,\,\,$ $\alpha \beta \,=\, \dfrac{c}{a}$

Now, the general form quadratic equation $a\bigg(x^2+\Big(\dfrac{b}{a}\Big)x+\dfrac{c}{a}\bigg) \,=\, 0$ can be expressed in terms of sum and product of the zeros (or roots).

$\implies$ $a\bigg(x^2-\Big(-\dfrac{b}{a}\Big)x+\dfrac{c}{a}\bigg) \,=\, 0$

$\implies$ $a\Big(x^2-(\alpha+\beta)x+\alpha \beta\Big)$ $\,=\,$ $0$

Now, it is time to factor this quadratic equation mathematically.

$\implies$ $a\Big(x^2-x \times (\alpha+\beta)+\alpha \times \beta\Big)$ $\,=\,$ $0$

$\implies$ $a(x^2-x \times \alpha$ $-$ $x \times \beta+\beta \times \alpha)$ $\,=\,$ $0$

$\implies$ $a\Big(x \times x-x \times \alpha$ $-$ $\beta \times x-\beta \times (-\alpha)\Big)$ $\,=\,$ $0$

$\implies$ $a\Big(x \times (x-\alpha)$ $-$ $\beta \times (x-\alpha)\Big)$ $\,=\,$ $0$

$\implies$ $a\Big((x-\alpha) \times x$ $-$ $(x-\alpha) \times \beta\Big)$ $\,=\,$ $0$

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

Therefore, the quadratic equation $ax^2+bx+c = 0$ is factored as $a(x-\alpha)(x-\beta) = 0$

List of the steps for factoring a quadratic equation and an example to learn how to solve the quadratic equation by the factorization (or factorisation) method.

List of the questions on quadratic equations with step by step solution to learn how to factor a quadratic equation for solving the quadratic equation by factoring.

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved