Solution - Factoring binomials using the difference of squares

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "0.81" was replaced by "(81/100)".

Rearrange:

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

                     p^2-((81/100))=0 

Step by step solution :

Step  1  :

             81
 Simplify   ———
            100

Equation at the end of step  1  :

           81
  (p2) -  ———  = 0 
          100
 Step  2  :

Rewriting the whole as an Equivalent Fraction :

 2.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  100  as the denominator :

           p2     p2 • 100
     p2 =  ——  =  ————————
           1        100   

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 2.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 p2 • 100 - (81)     100p2 - 81
 ———————————————  =  ——————————
       100              100    

Trying to factor as a Difference of Squares :

 2.3      Factoring:  100p² - 81 

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 :  100  is the square of  10 
Check : 81 is the square of 9
Check :  p²  is the square of  p¹ 

Factorization is :       (10p + 9)  •  (10p - 9) 

Equation at the end of step  2  :

  (10p + 9) • (10p - 9)
  —————————————————————  = 0 
           100         

Step  3  :

When a fraction equals zero :

 3.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  (10p+9)•(10p-9)
  ——————————————— • 100 = 0 • 100
        100      

Now, on the left hand side, the  100  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   (10p+9)  •  (10p-9)  = 0

Theory - Roots of a product :

 3.2    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 :

 3.3      Solve  :    10p+9 = 0 

 Subtract  9  from both sides of the equation : 
                      10p = -9
Divide both sides of the equation by 10:
                     p = -9/10 = -0.900

Solving a Single Variable Equation :

 3.4      Solve  :    10p-9 = 0 

 Add  9  to both sides of the equation : 
                      10p = 9
Divide both sides of the equation by 10:
                     p = 9/10 = 0.900

Two solutions were found :

1.  p = 9/10 = 0.900
   
2.  p = -9/10 = -0.900