Solution - Adding, subtracting and finding the least common multiple

Step  1  :

              5  
 Simplify   —————
            x + 3

Equation at the end of step  1  :

      2          3          5  
  (——————— +  ———————) +  —————
   (x + 1)    (x + 2)     x + 3

Step  2  :

              3  
 Simplify   —————
            x + 2

Equation at the end of step  2  :

      2         3         5  
  (——————— +  —————) +  —————
   (x + 1)    x + 2     x + 3

Step  3  :

              2  
 Simplify   —————
            x + 1

Equation at the end of step  3  :

     2        3         5  
  (————— +  —————) +  —————
   x + 1    x + 2     x + 3

Step  4  :

Calculating the Least Common Multiple :

 4.1    Find the Least Common Multiple

      The left denominator is :       x+1 

      The right denominator is :       x+2 

      Least Common Multiple:
      (x+1) • (x+2) 

Calculating Multipliers :

 4.2    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = x+2

   Right_M = L.C.M / R_Deno = x+1

Making Equivalent Fractions :

 4.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)²   and  (y²+y)/(y+1)³  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

   L. Mult. • L. Num.        2 • (x+2)  
   ——————————————————  =   —————————————
         L.C.M             (x+1) • (x+2)

   R. Mult. • R. Num.        3 • (x+1)  
   ——————————————————  =   —————————————
         L.C.M             (x+1) • (x+2)

Adding fractions that have a common denominator :

 4.4       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:

 2 • (x+2) + 3 • (x+1)           5x + 7     
 —————————————————————  =  —————————————————
     (x+1) • (x+2)         (x + 1) • (x + 2)

Equation at the end of step  4  :

       (5x + 7)          5  
  ————————————————— +  —————
  (x + 1) • (x + 2)    x + 3

Step  5  :

Calculating the Least Common Multiple :

 5.1    Find the Least Common Multiple

      The left denominator is :       (x+1) • (x+2) 

      The right denominator is :       x+3 

      Least Common Multiple:
      (x+1) • (x+2) • (x+3) 

Calculating Multipliers :

 5.2    Calculate multipliers for the two fractions


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = x+3

   Right_M = L.C.M / R_Deno = (x+1)•(x+2)

Making Equivalent Fractions :

 5.3      Rewrite the two fractions into equivalent fractions

   L. Mult. • L. Num.          (5x+7) • (x+3)   
   ——————————————————  =   —————————————————————
         L.C.M             (x+1) • (x+2) • (x+3)

   R. Mult. • R. Num.        5 • (x+1) • (x+2)  
   ——————————————————  =   —————————————————————
         L.C.M             (x+1) • (x+2) • (x+3)

Adding fractions that have a common denominator :

 5.4       Adding up the two equivalent fractions

 (5x+7) • (x+3) + 5 • (x+1) • (x+2)           10x2 + 37x + 31      
 ——————————————————————————————————  =  ———————————————————————————
       (x+1) • (x+2) • (x+3)            (x + 1) • (x + 2) • (x + 3)

Trying to factor by splitting the middle term

 5.5     Factoring  10x² + 37x + 31 

The first term is,  10x²  its coefficient is  10 .
The middle term is,  +37x  its coefficient is  37 .
The last term, "the constant", is  +31 

Step-1 : Multiply the coefficient of the first term by the constant   10 • 31 = 310 

Step-2 : Find two factors of  310  whose sum equals the coefficient of the middle term, which is   37 .

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Final result :

        10x2 + 37x + 31      
  ———————————————————————————
  (x + 1) • (x + 2) • (x + 3)