The real number system

Question

please how do I solve the equation in a real number system 

 

4x^3-19x^2+19x+6=0

Answer 1

f(x) = 4x^3-19x^2+19x+6

Since f(3) = 0, x-3 is a factor. By synthetic division,

3|4 -19 19 6

......12 -21 6

------------------

...4 -7 -2 | 0

So, another factor is 4x^2-7x-2.

Solve 4x^2-7x-2 = (4x+1)(x-2) = 0

x = -1/4, 2

Answer: x = -1/4, 2, 3

Answer 2

Let see if x = ±2, ±3, ±6 are root, because these are the simplest roots to test for.
4*2³ - 19*2² + 19*2 + 6 = 0  ⇒  x₁ = 2

Let's divide given polynomial by (x - 2)

      
             4x² - 11x - 3             
(x - 2) | 4x³ - 19x² + 19x + 6
           -
             4x³ - 8x²            
                   - 11x²  + 19x
                 -    
                    - 11x² + 22x      
                                 - 3x + 6
                              -
                                 - 3x + 6   
                                        0
   
(x - 2)(4x² - 11x - 3) = 0

4x² - 11x - 3 = 0
D = b² - 4ac = (-11)² - 4*4*(-3) = 169

          - b ± √D
x₂₃ = -------------
              2a 

          -(-11) ± √169
x₂₃ = ------------------- 
                  2*4


x₂ = (11 + 13) / 8 = 3
x₃ = (11 - 13) / 8 = - 2/8 = - 1/4 

The roots of a given equation are {- 1/4, 2, 3} .