Simplify: (2x-1)^3=-125 Tiger Algebra Solver

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     (2*x-1)^3-(-125)=0 

 2.1     Pull out like factors :

   8x³ - 12x² + 6x + 124  = 

  2 • (4x³ - 6x² + 3x + 62) 

 2.3      Factoring:  4x³ - 6x² + 3x + 62 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  3x + 62 
Group 2:  -6x² + 4x³ 

Pull out from each group separately :

Group 1:   (3x + 62) • (1)
Group 2:   (2x - 3) • (2x²)

Bad news !! Factoring by pulling out fails : The groups have no common factor and can not be added up to form a multiplication.

 2.4    Find roots (zeroes) of :       F(x) = 4x³ - 6x² + 3x + 62
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  4  and the Trailing Constant is  62. The factor(s) are:

of the Leading Coefficient :  1,2 ,4
 of the Trailing Constant :  1 ,2 ,31 ,62 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
   4x³ - 6x² + 3x + 62 
can be divided with  x + 2 

 2.6     Factoring  4x²-14x+31 

The first term is,  4x²  its coefficient is  4 .
The middle term is,  -14x  its coefficient is  -14 .
The last term, "the constant", is  +31 

Step-1 : Multiply the coefficient of the first term by the constant   4 • 31 = 124 

Step-2 : Find two factors of  124  whose sum equals the coefficient of the middle term, which is   -14 .

 3.2      Solve :    2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Root plot for :  y = 4x²-14x+31
Axis of Symmetry (dashed)  {x}={ 1.75} 
Vertex at  {x,y} = { 1.75,18.75} 
Function has no real roots

 3.4     Solving   4x²-14x+31 = 0 by Completing The Square .Divide both sides of the equation by  4  to have 1 as the coefficient of the first term :
   x²-(7/2)x+(31/4) = 0

Subtract  31/4  from both side of the equation :
   x²-(7/2)x = -31/4

Now the clever bit: Take the coefficient of  x , which is  7/2 , divide by two, giving  7/4 , and finally square it giving  49/16 

Add  49/16  to both sides of the equation :
  On the right hand side we have :
   -31/4  +  49/16   The common denominator of the two fractions is  16   Adding  (-124/16)+(49/16)  gives  -75/16 
  So adding to both sides we finally get :
   x²-(7/2)x+(49/16) = -75/16

Adding  49/16  has completed the left hand side into a perfect square :
   x²-(7/2)x+(49/16)  =
   (x-(7/4)) • (x-(7/4))  =
  (x-(7/4))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(7/2)x+(49/16) = -75/16 and
   x²-(7/2)x+(49/16) = (x-(7/4))²
then, according to the law of transitivity,
   (x-(7/4))² = -75/16

We'll refer to this Equation as  Eq. #3.4.1  

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

Note that the square root of
   (x-(7/4))²   is
   (x-(7/4))^2/2 =
  (x-(7/4))¹ =
   x-(7/4)

Now, applying the Square Root Principle to  Eq. #3.4.1  we get:
   x-(7/4) = √ -75/16

Add  7/4  to both sides to obtain:
   x = 7/4 + √ -75/16
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 - (7/2)x + (31/4) = 0
   has two solutions:
  x = 7/4 + √ 75/16 •  i 
   or
  x = 7/4 - √ 75/16 •  i 

Note that  √ 75/16 can be written as
  √ 75  / √ 16   which is √ 75  / 4

 3.5     Solving    4x²-14x+31 = 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   =     4
                      B   =   -14
                      C   =   31 Accordingly,  B²  -  4AC   =
                     196 - 496 =
                     -300Applying the quadratic formula :

               14 ± √ -300
   x  =    ——————
                      8In 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,√ -300  = 
                    √ 300 • (-1)  =
                    √ 300  • √ -1   =
                    ±  √ 300  • i

Can  √ 300 be simplified ?

Yes!   The prime factorization of  300   is
   2•2•3•5•5 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 300   =  √ 2•2•3•5•5   =2•5•√ 3   =
                ±  10 • √ 3

  √ 3   , rounded to 4 decimal digits, is   1.7321
 So now we are looking at:
           x  =  ( 14 ± 10 •  1.732 i ) / 8

Two imaginary solutions :