Solution - Finding the roots of polynomials

Step  1  :

Equation at the end of step  1  :

                     7
  (((((3•(a2))+2a)-(——•(a4)))+(2•32a2))-2a)+42
                    27
 Step  2  :
             7
 Simplify   ——
            27

Equation at the end of step  2  :

                     7
  (((((3•(a2))+2a)-(——•a4))+(2•32a2))-2a)+42
                    27

Step  3  :

Equation at the end of step  3  :

                   7a4
  (((((3•(a2))+2a)-———)+(2•32a2))-2a)+42
                   27 
 Step  4  :

Equation at the end of step  4  :

                    7a4                           
  ((((3a2 +  2a) -  ———) +  (2•32a2)) -  2a) +  42
                    27                            

Step  5  :

Rewriting the whole as an Equivalent Fraction :

 5.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  27  as the denominator :

                 3a2 + 2a     (3a2 + 2a) • 27
     3a2 + 2a =  ————————  =  ———————————————
                    1               27       

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

Step  6  :

Pulling out like terms :

 6.1     Pull out like factors :

   3a² + 2a  =   a • (3a + 2) 

Adding fractions that have a common denominator :

 6.2       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:

 a • (3a+2) • 27 - (7a4)     -7a4 + 81a2 + 54a
 ———————————————————————  =  —————————————————
           27                       27        

Equation at the end of step  6  :

    (-7a4 + 81a2 + 54a)                          
  ((——————————————————— +  (2•32a2)) -  2a) +  42
            27                                   

Step  7  :

Rewriting the whole as an Equivalent Fraction :

 7.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  27  as the denominator :

                (2•32a2)     (2•32a2) • 27
    (2•32a2) =  ————————  =  —————————————
                   1              27      

Step  8  :

Pulling out like terms :

 8.1     Pull out like factors :

   -7a⁴ + 81a² + 54a  =   -a • (7a³ - 81a - 54) 

Polynomial Roots Calculator :

 8.2    Find roots (zeroes) of :       F(a) = 7a³ - 81a - 54
Polynomial Roots Calculator is a set of methods aimed at finding values of  a  for which   F(a)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  a  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  7  and the Trailing Constant is  -54.

 The factor(s) are:

of the Leading Coefficient :  1,7
 of the Trailing Constant :  1 ,2 ,3 ,6 ,9 ,18 ,27 ,54

 Let us test ....

Note - For tidiness, printing of 27 checks which found no root was suppressed

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
   7a³ - 81a - 54 
can be divided with  a + 3 

Polynomial Long Division :

 8.3    Polynomial Long Division
Dividing :  7a³ - 81a - 54 
                              ("Dividend")
By         :    a + 3    ("Divisor")

Quotient :  7a²-21a-18  Remainder:  0 

Trying to factor by splitting the middle term

 8.4     Factoring  7a²-21a-18 

The first term is,  7a²  its coefficient is  7 .
The middle term is,  -21a  its coefficient is  -21 .
The last term, "the constant", is  -18 

Step-1 : Multiply the coefficient of the first term by the constant   7 • -18 = -126 

Step-2 : Find two factors of  -126  whose sum equals the coefficient of the middle term, which is   -21 .

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

Adding fractions that have a common denominator :

 8.5       Adding up the two equivalent fractions

 -a • (7a2-21a-18) • (a+3) + (2•32a2) • 27      -7a4 + 567a2 + 54a
 —————————————————————————————————————————  =  ——————————————————
                    27                                 27        

Equation at the end of step  8  :

   (-7a4 + 567a2 + 54a)           
  (———————————————————— -  2a) +  42
            27                    

Step  9  :

Rewriting the whole as an Equivalent Fraction :

 9.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  27  as the denominator :

          2a     2a • 27
    2a =  ——  =  ———————
          1        27   

Step  10  :

Pulling out like terms :

 10.1     Pull out like factors :

   -7a⁴ + 567a² + 54a  =   -a • (7a³ - 567a - 54) 

Polynomial Roots Calculator :

 10.2    Find roots (zeroes) of :       F(a) = 7a³ - 567a - 54

     See theory in step 8.2
In this case, the Leading Coefficient is  7  and the Trailing Constant is  -54.

 The factor(s) are:

of the Leading Coefficient :  1,7
 of the Trailing Constant :  1 ,2 ,3 ,6 ,9 ,18 ,27 ,54

 Let us test ....

Note - For tidiness, printing of 27 checks which found no root was suppressed

Polynomial Roots Calculator found no rational roots

Adding fractions that have a common denominator :

 10.3       Adding up the two equivalent fractions

 -a • (7a3-567a-54) - (2a • 27)     567a2 - 7a4
 ——————————————————————————————  =  ———————————
               27                       27     

Equation at the end of step  10  :

  (567a2 - 7a4)    
  ————————————— +  42
       27          

Step  11  :

Rewriting the whole as an Equivalent Fraction :

 11.1   Adding a whole to a fraction

Rewrite the whole as a fraction using  27  as the denominator :

          42     42 • 27
    42 =  ——  =  ———————
          1        27   

Step  12  :

Pulling out like terms :

 12.1     Pull out like factors :

   567a² - 7a⁴  =   -7a² • (a² - 81) 

Trying to factor as a Difference of Squares :

 12.2      Factoring:  a² - 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 : 81 is the square of 9
Check :  a²  is the square of  a¹ 

Factorization is :       (a + 9)  •  (a - 9) 

Adding fractions that have a common denominator :

 12.3       Adding up the two equivalent fractions

 -7a2 • (a+9) • (a-9) + 42 • 27     -7a4 + 567a2 + 1134
 ——————————————————————————————  =  ———————————————————
               27                           27         

Step  13  :

Pulling out like terms :

 13.1     Pull out like factors :

   -7a⁴ + 567a² + 1134  =   -7 • (a⁴ - 81a² - 162) 

Trying to factor by splitting the middle term

 13.2     Factoring  a⁴ - 81a² - 162 

The first term is,  a⁴  its coefficient is  1 .
The middle term is,  -81a²  its coefficient is  -81 .
The last term, "the constant", is  -162 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -162 = -162 

Step-2 : Find two factors of  -162  whose sum equals the coefficient of the middle term, which is   -81 .

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

Final result :

  -7 • (a4 - 81a2 - 162)
  ——————————————————————
            27