Solution - Factoring binomials using the difference of squares

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

           (6*m^2-1)*(6*m^2-1)-(a*m^4+b*m^2+1)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((6•(m2))-1)•((2•3m2)-1))-(m4a+m2b+1)  = 0 
 Step  2  :

Equation at the end of step  2  :

  (((2•3m2) -  1) • (6m2 - 1)) -  (m4a + m2b + 1)  = 0 

Step  3  :

Trying to factor as a Difference of Squares :

 3.1      Factoring:  6m²-1 

Theory : A difference of two perfect squares,  A² - B²  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A² - AB + BA - B² =
         A² - AB + AB - B² =
         A² - B²

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check :  6  is not a square !!

Ruling : Binomial can not be factored as the
difference of two perfect squares

Trying to factor as a Difference of Squares :

 3.2      Factoring:  6m²-1 

Check :  6  is not a square !!

Ruling : Binomial can not be factored as the
difference of two perfect squares

Evaluate an expression :

 3.3    Multiply  (6m²-1)  by  (6m²-1) 

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (6m²-1)  and the exponents are :
          1 , as  (6m²-1)  is the same number as  (6m²-1)¹ 
 and   1 , as  (6m²-1)  is the same number as  (6m²-1)¹ 
The product is therefore,  (6m²-1)⁽¹⁺¹⁾ = (6m²-1)² 

Equation at the end of step  3  :

  (6m2 - 1)2 -  (m4a + m2b + 1)  = 0 

Step  4  :

 4.1     Evaluate :  (6m2-1)2   =    36m4-12m2+1 

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   -m⁴a + 36m⁴ - m²b - 12m²  = 

  -m² • (m²a - 36m² + b + 12) 

Equation at the end of step  5  :

  -m2 • (m2a - 36m2 + b + 12)  = 0 

Step  6  :

Theory - Roots of a product :

 6.1    A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 6.2      Solve  :    -m² = 0 

 Multiply both sides of the equation by (-1) :  m² = 0


 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      m  =  ± √ 0  

 Any root of zero is zero. This equation has one solution which is  m = 0

Solving a Single Variable Equation :

 6.3     Solve   m²a-36m²+b+12  = 0

In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.

We shall not handle this type of equations at this time.

One solution was found :