Solution - Factoring binomials using the difference of squares

Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "x2"   was replaced by   "x^2".  4 more similar replacement(s).

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  ((((((5•(x6))-(x5))-(5•(x4)))+(2•3x3))-x2)-5x)+1  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((((((5•(x6))-(x5))-5x4)+(2•3x3))-x2)-5x)+1  = 0 
 Step  3  :

Equation at the end of step  3  :

  (((((5x6-x5)-5x4)+(2•3x3))-x2)-5x)+1  = 0 

Step  4  :

Polynomial Roots Calculator :

 4.1    Find roots (zeroes) of :       F(x) = 5x⁶-x⁵-5x⁴+6x³-x²-5x+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  5  and the Trailing Constant is  1.

 The factor(s) are:

of the Leading Coefficient :  1,5
 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
   5x⁶-x⁵-5x⁴+6x³-x²-5x+1 
can be divided by 3 different polynomials,including by  5x-1 

Polynomial Long Division :

 4.2    Polynomial Long Division
Dividing :  5x⁶-x⁵-5x⁴+6x³-x²-5x+1 
                              ("Dividend")
By         :    5x-1    ("Divisor")

Quotient :  x⁵-x³+x²-1  Remainder:  0 

Polynomial Roots Calculator :

 4.3    Find roots (zeroes) of :       F(x) = x⁵-x³+x²-1

     See theory in step 4.1
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
   x⁵-x³+x²-1 
can be divided by 2 different polynomials,including by  x-1 

Polynomial Long Division :

 4.4    Polynomial Long Division
Dividing :  x⁵-x³+x²-1 
                              ("Dividend")
By         :    x-1    ("Divisor")

Quotient :  x⁴+x³+x+1  Remainder:  0 

Polynomial Roots Calculator :

 4.5    Find roots (zeroes) of :       F(x) = x⁴+x³+x+1

     See theory in step 4.1
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
   x⁴+x³+x+1 
can be divided with  x+1 

Polynomial Long Division :

 4.6    Polynomial Long Division
Dividing :  x⁴+x³+x+1 
                              ("Dividend")
By         :    x+1    ("Divisor")

Quotient :  x³+1  Remainder:  0 

Trying to factor as a Sum of Cubes :

 4.7      Factoring:  x³+1 

Theory : A sum of two perfect cubes,  a³ + b³ can be factored into  :
             (a+b) • (a²-ab+b²)
Proof  : (a+b) • (a²-ab+b²) =
    a³-a²b+ab²+ba²-b²a+b³ =
    a³+(a²b-ba²)+(ab²-b²a)+b³=
    a³+0+0+b³=
    a³+b³

Check :  1  is the cube of   1 
Check :  x³ is the cube of   x¹

Factorization is :
             (x + 1)  •  (x² - x + 1) 

Trying to factor by splitting the middle term

 4.8     Factoring  x² - x + 1 

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

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

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -1 .

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

Multiplying Exponential Expressions :

 4.9    Multiply  (x+1)  by  (x+1) 

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

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

Equation at the end of step  4  :

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

Step  5  :

Theory - Roots of a product :

 5.1    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 :

 5.2      Solve  :    (x+1)² = 0 

  (x+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 :   x+1  = 0

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

Parabola, Finding the Vertex :

 5.3      Find the Vertex of   y = x²-x+1

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   0.5000  

 Plugging into the parabola formula   0.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * 0.50 * 0.50 - 1.0 * 0.50 + 1.0
or   y = 0.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x²-x+1
Axis of Symmetry (dashed)  {x}={ 0.50} 
Vertex at  {x,y} = { 0.50, 0.75} 
Function has no real roots

Solve Quadratic Equation by Completing The Square

 5.4     Solving   x²-x+1 = 0 by Completing The Square .

 Subtract  1  from both side of the equation :
   x²-x = -1

Now the clever bit: Take the coefficient of  x , which is  1 , divide by two, giving  1/2 , and finally square it giving  1/4 

Add  1/4  to both sides of the equation :
  On the right hand side we have :
   -1  +  1/4    or,  (-1/1)+(1/4) 
  The common denominator of the two fractions is  4   Adding  (-4/4)+(1/4)  gives  -3/4 
  So adding to both sides we finally get :
   x²-x+(1/4) = -3/4

Adding  1/4  has completed the left hand side into a perfect square :
   x²-x+(1/4)  =
   (x-(1/2)) • (x-(1/2))  =
  (x-(1/2))²
Things which are equal to the same thing are also equal to one another. Since
   x²-x+(1/4) = -3/4 and
   x²-x+(1/4) = (x-(1/2))²
then, according to the law of transitivity,
   (x-(1/2))² = -3/4

We'll refer to this Equation as  Eq. #5.4.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x-(1/2))²   is
   (x-(1/2))^2/2 =
  (x-(1/2))¹ =
   x-(1/2)

Now, applying the Square Root Principle to  Eq. #5.4.1  we get:
   x-(1/2) = √ -3/4

Add  1/2  to both sides to obtain:
   x = 1/2 + √ -3/4
In Math,  i  is called the imaginary unit. It satisfies   i²  =-1. Both   i   and   -i   are the square roots of   -1 


Since a square root has two values, one positive and the other negative
   x² - x + 1 = 0
   has two solutions:
  x = 1/2 + √ 3/4 •  i 
   or
  x = 1/2 - √ 3/4 •  i 

Note that  √ 3/4 can be written as
  √ 3  / √ 4   which is √ 3  / 2

Solve Quadratic Equation using the Quadratic Formula

 5.5     Solving    x²-x+1 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax²+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  x =   ————————
                      2A

  In our case,  A   =     1
                      B   =    -1
                      C   =   1

Accordingly,  B²  -  4AC   =
                     1 - 4 =
                     -3

Applying the quadratic formula :

               1 ± √ -3
   x  =    —————
                    2

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -3  = 
                    √ 3 • (-1)  =
                    √ 3  • √ -1   =
                    ±  √ 3  • i

  √ 3   , rounded to 4 decimal digits, is   1.7321
 So now we are looking at:
           x  =  ( 1 ±  1.732 i ) / 2

Two imaginary solutions :

 x =(1+√-3)/2=(1+i√ 3 )/2= 0.5000+0.8660i
  or: 
 x =(1-√-3)/2=(1-i√ 3 )/2= 0.5000-0.8660i

Solving a Single Variable Equation :

 5.6      Solve  :    x-1 = 0 

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

Solving a Single Variable Equation :

 5.7      Solve  :    5x-1 = 0 

 Add  1  to both sides of the equation : 
                      5x = 1
Divide both sides of the equation by 5:
                     x = 1/5 = 0.200

5 solutions were found :

1.  x = 1/5 = 0.200
   
2.  x = 1
3.  x =(1-√-3)/2=(1-i√ 3 )/2= 0.5000-0.8660i
4.  x =(1+√-3)/2=(1+i√ 3 )/2= 0.5000+0.8660i
5.  x = -1