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 x4-10x2+9=0 Four solutions were found : x = 3 x = -3 x = 1 x = -1

Step by step solution :Step  1  :Skip Ad
Equation at the end of step  1  : ((x4) - (2•5x2)) + 9 = 0 Step  2  :Trying to factor by splitting the middle term

 2.1     Factoring  x4-10x2+9 

The first term is,  x4  its coefficient is  1 .
The middle term is,  -10x2  its coefficient is  -10 .
The last term, "the constant", is   +9 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 9 = 9 

Step-2 : Find two factors of   9  whose sum equals the coefficient of the middle term, which is   -10 .

     -9   +   -1   =   -10   That's it


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  -1 
                     x4 - 9x2 - 1x2 - 9

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x2 • (x2-9)
              Add up the last 2 terms, pulling out common factors :
                     1 • (x2-9)
Step-5 : Add up the four terms of step 4 :
                    (x2-1)  •  (x2-9)
             Which is the desired factorization

Trying to factor as a Difference of Squares : 

 2.2      Factoring:  x2-1 

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 : 1 is the square of 1
Check :  x2  is the square of   x1 

Factorization is :       (x + 1)  •  (x - 1) 

Trying to factor as a Difference of Squares : 

 2.3      Factoring:  x2 - 9 

Check : 9 is the square of 3
Check :  x2  is the square of   x1 

Factorization is :       (x + 3)  •  (x - 3) 

Equation at the end of step  2  : (x + 1) • (x - 1) • (x + 3) • (x - 3) = 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  :    x+1 = 0 

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

Solving a Single Variable Equation : 

 3.3      Solve  :    x-1 = 0 

 Add  1  to both sides of the equation : 
                      x = 1 

Solving a Single Variable Equation : 

 3.4      Solve  :    x+3 = 0 

 Subtract  3  from both sides of the equation : 
                      x = -3 

Solving a Single Variable Equation : 

 3.5      Solve  :    x-3 = 0 

 Add  3  to both sides of the equation : 
                      x = 3 

Supplement : Solving Quadratic Equation DirectlySolving  x4-10x2+9  = 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   x4-10x2+9 = 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 :
 w2-10w+9 = 0

Solving this new equation using the quadratic formula we get two real solutions :
   9.0000  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 
   x4-10x2+9 = 0
  are either : 
  x =√ 9.000 = 3.00000  or :
  x =√ 9.000 = -3.00000  or :
  x =√ 1.000 = 1.00000  or :
  x =√ 1.000 = -1.00000 

Four solutions were found : x = 3 x = -3 x = 1 x = -1


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