Solving Equations by Factoring

Introduction to solving equations by factoring:
What is an Equation ?

An equation is a mathematical representation of two expressions which are equal to each other. Every equation has two parts one is the

Left Hand Side and other is the Right Hand Side. The left hand side is connected to the right hand side by an equal sign.

Properties of an Equation

1) Any quantity can be added to both the sides of an equation.

2) Any quantity can be subtracted to both the sides of an equation.

3) Any quantity can be multiplied to both the sides of an equation.

4) Any non zero quantity can divide both the sides of an equation.

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Solving Equations by Factoring - Different Methods

Equations can be solved in many ways.

1) By means of factoring.

2) By taking the roots.

3) By completing the square in case of quadratic equations.

4) By using the quadratic formula.

5) By using the method of graphing.

In this article we will discuss solving equations by means of factoring.

In this context lets first have the idea of ' Zero Factor Principle ' .

What is Zero factor principle ?

Zero factor principle states that product of two expressions is zero , if and only if either of the expression is zero.

Explanation : Let A and B be two expressions.

Product of A and B is AB.

AB = 0 , the relation holds true if and only if either A = 0  or B = 0 .

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Solving Equations by Factoring-examples

Now lets see some examples.

d  a = 5 / 3.

Examples :

1 )    x ^2   -  5x   -  6  = 0

Or,    x^2 - 6x + x - 6  = 0    [ Always break the middle term , such that even after breaking there is no change in the expression ]

Or,    x ( x - 6 ) + 1 ( x - 6 ) = 0

Or,    ( x - 6 ) ( x + 1) = 0

Now from Zero factor principle we can conclude that either ( x - 6 ) = 0   or  ( x + 1 ) = 0

When ( x - 6 ) = 0                                           When ( x + 1) = 0

Or,       x   =   6                                                           Or, x = - 1

Therefore the equation has two roots x = 6  and x = - 1

Explanation

Procedure to break the middle term.

First we will find the product of the coefficient of x^2 and the constant term in the equation.

In this case it is 1 * ( - 6 ) = -6

Now we will find two numbers that multiply to give -6 ( product of the coefficient of x^2 and the constant term )

and add to give -5 ( coefficient of x ).

The probable numbers are

a)  2 and -3 :     2 * ( - 3) = -6   but 2 + ( - 3 ) = -1 so it will not be accepted.

b)  - 2 and 3 :    ( - 2 ) *  3 = -6   but ( - 2 ) +  3  = 1 so it will not be accepted.

c)  6 and -1 :     6 * ( - 1) = -6   but 6 + ( - 1 ) = 5 so it will not be accepted.

d)  -6 and 1 :     1 * ( - 6 ) = -6   but 1 + ( - 6 ) = - 5 so it will  be accepted.

So that is why the middle term is represented as 5x  = - 6x + x .

2)     x^2   -  7x   +  12  =  0

Or,   x^2  -  4x   -  3x   +  12 = 0       [ Always break the middle term , such that even after breaking there is no change in the expression ]

Or ,   x ( x - 4) - 3 ( x  - 4 ) = 0

Or,    ( x - 4 ) ( x - 3 ) =  0

Now from Zero factor principle we can conclude that either ( x - 4 ) = 0    or   ( x - 3 ) =  0

When   ( x - 4 ) = 0                                                     When   ( x - 3 ) = 0         

Or,        x  =  4                                                                        Or,  x  =  3

Therefore the equation has two roots x = 4  and x = 3

Explanation

Procedure to break the middle term.

First we will find the product of the coefficient of x^2 and the constant term in the equation.

In this case it is 1 *  12  =  12

Now we will find two numbers that multiply to give  12( product of the coefficient of x^2 and the constant term )

and add to give -7 ( coefficient of x ).

The numbers are   - 4   and  - 3 :

( - 4 ) * ( - 3 ) = 12   and   ( - 4 ) + ( - 3 )   =   - 7   so it will be accepted.

So that is why the middle term is represented as   - 7 x  =   - 4x   -  3 x .

3)     6a^4  -  13a^3  + 5a^2  =  0

Or,    a^2 ( 6a^2 - 13a + 5 ) = 0

Or,    a^2 ( 6a^2 - 10a  -  3a  +  5 ) = 0

Or,    a^2 { 2a ( 3a - 5 ) - 1 ( 3a - 5 ) } = 0

Or,    a^2 ( 2a - 1 ) ( 3a - 5 )= 0

Now from Zero factor principle we can conclude that  a^2 = 0   or  ( 2a - 1 ) = 0    or   ( 3a - 5 ) =  0

When ( 2a - 1 ) = 0                                                   When ( 3a - 5 ) = 0                                  When a^2 = 0

Or,        2a = 1                                                                Or,       3a  =  5                                        Or, a = 0 

Or,          a =  1 / 2                                                           Or,        a = 5 / 3

Therefore the equation has  roots as a = 0  ,  a = 1 / 2  an