Solution - Equations reducible to quadratic form

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((((x2) +  1)2) -  5x2) -  5  = 0 

Step  2  :

Trying to factor by splitting the middle term

 2.1     Factoring  x⁴-3x²-4 

The first term is,  x⁴  its coefficient is  1 .
The middle term is,  -3x²  its coefficient is  -3 .
The last term, "the constant", is  -4 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -4 = -4 

Step-2 : Find two factors of  -4  whose sum equals the coefficient of the middle term, which is   -3 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -4  and  1 
                     x⁴ - 4x² + 1x² - 4

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x² • (x²-4)
              Add up the last 2 terms, pulling out common factors :
                     1 • (x²-4)
Step-5 : Add up the four terms of step 4 :
                    (x²+1)  •  (x²-4)
             Which is the desired factorization

Polynomial Roots Calculator :

 2.2    Find roots (zeroes) of :       F(x) = x²+1
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  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  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

Trying to factor as a Difference of Squares :

 2.3      Factoring:  x²-4 

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 : 4 is the square of 2
Check :  x²  is the square of  x¹ 

Factorization is :       (x + 2)  •  (x - 2) 

Equation at the end of step  2  :

  (x2 + 1) • (x + 2) • (x - 2)  = 0 

Step  3  :

Theory - Roots of a product :

 3.1    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 :

 3.2      Solve  :    x²+1 = 0 

 Subtract  1  from both sides of the equation : 
                      x² = -1
 
 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  
                      x  =  ± √ -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.

                      x=  0.0000 + 1.0000 i 
                      x=  0.0000 - 1.0000 i 

Solving a Single Variable Equation :

 3.3      Solve  :    x+2 = 0 

 Subtract  2  from both sides of the equation : 
                      x = -2

Solving a Single Variable Equation :

 3.4      Solve  :    x-2 = 0 

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

Supplement : Solving Quadratic Equation Directly

Solving    x4-3x2-4  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

 4.1     Solve   x⁴-3x²-4 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that  w = x²  transforms the equation into :
 w²-3w-4 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   4.0000  or  -1.0000

Now that we know the value(s) of  w , we can calculate  x  since  x  is  √ w  

Doing just this we discover that the solutions of
   x⁴-3x²-4 = 0
  are either : 
  x =√ 4.000 = 2.00000  or :
  x =√ 4.000 = -2.00000  or :
  x =√-1.000 = 0.0 + 1.00000 i  or :
  x =√-1.000 = 0.0 - 1.00000 i

Four solutions were found :

1.  x = 2
2.  x = -2
3.   x=  0.0000 - 1.0000 i 
   
4.   x=  0.0000 + 1.0000 i