Solution - Polynomial long division

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "d2"   was replaced by   "d^2".  2 more similar replacement(s).

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((((d5)-(2•(d3)))-2d2)-3d)-2)•y  = 0 
 Step  2  :

Equation at the end of step  2  :

  (((((d5)-2d3)-2d2)-3d)-2)•y  = 0 

Step  3  :

Polynomial Roots Calculator :

 3.1    Find roots (zeroes) of :       F(d) = d⁵-2d³-2d²-3d-2
Polynomial Roots Calculator is a set of methods aimed at finding values of  d  for which   F(d)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  d  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  1  and the Trailing Constant is  -2.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1 ,2

 Let us test ....

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
   d⁵-2d³-2d²-3d-2 
can be divided by 2 different polynomials,including by  d-2 

Polynomial Long Division :

 3.2    Polynomial Long Division
Dividing :  d⁵-2d³-2d²-3d-2 
                              ("Dividend")
By         :    d-2    ("Divisor")

Quotient :  d⁴+2d³+2d²+2d+1  Remainder:  0 

Polynomial Roots Calculator :

 3.3    Find roots (zeroes) of :       F(d) = d⁴+2d³+2d²+2d+1

     See theory in step 3.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  1.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1

 Let us test ....

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
   d⁴+2d³+2d²+2d+1 
can be divided with  d+1 

Polynomial Long Division :

 3.4    Polynomial Long Division
Dividing :  d⁴+2d³+2d²+2d+1 
                              ("Dividend")
By         :    d+1    ("Divisor")

Quotient :  d³+d²+d+1  Remainder:  0 

Polynomial Roots Calculator :

 3.5    Find roots (zeroes) of :       F(d) = d³+d²+d+1

     See theory in step 3.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  1.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1

 Let us test ....

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
   d³+d²+d+1 
can be divided with  d+1 

Polynomial Long Division :

 3.6    Polynomial Long Division
Dividing :  d³+d²+d+1 
                              ("Dividend")
By         :    d+1    ("Divisor")

Quotient :  d²+1  Remainder:  0 

Polynomial Roots Calculator :

 3.7    Find roots (zeroes) of :       F(d) = d²+1

     See theory in step 3.1
In this case, the Leading Coefficient is  1  and the Trailing Constant is  1.

 The factor(s) are:

of the Leading Coefficient :  1
 of the Trailing Constant :  1

 Let us test ....

Polynomial Roots Calculator found no rational roots

Multiplying Exponential Expressions :

 3.8    Multiply  (d+1)  by  (d+1) 

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

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

Equation at the end of step  3  :

  y • (d2 + 1) • (d + 1)2 • (d - 2)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    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 :

 4.2      Solve  :    y = 0 

  Solution is  y = 0

Solving a Single Variable Equation :

 4.3      Solve  :    d²+1 = 0 

 Subtract  1  from both sides of the equation : 
                      d² = -1
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      d  =  ± √ -1  

 In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 
The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      d=  0.0000 + 1.0000 i 
                      d=  0.0000 - 1.0000 i 

Solving a Single Variable Equation :

 4.4      Solve  :    (d+1)² = 0 

  (d+1) ² represents, in effect, a product of 2 terms which is equal to zero

For the product to be zero, at least one of these terms must be zero. Since all these terms are equal to each other, it actually means :   d+1  = 0

Subtract  1  from both sides of the equation : 
                      d = -1

Solving a Single Variable Equation :

 4.5      Solve  :    d-2 = 0 

 Add  2  to both sides of the equation : 
                      d = 2

5 solutions were found :

1.  d = 2
2.  d = -1
3.   d=  0.0000 - 1.0000 i 
   
4.   d=  0.0000 + 1.0000 i 
   
5.  y = 0