Question

Factorize x^4-3x-2
Thanks

Answer 1

P(x) = x⁴  - 3 x - 2
        = (x² + a x - b) ( x² + c x + 2 / b)             let  be....
        = x⁴ + (a+c) x³ + (-b+ a c +2/b) x²  + (-bc + 2a/b) x - 2 = 0
compare coefficients:
a = - c
b c - 2 a/b = 3   =>  b² c + 2 c - 3 b = 0          ---(1)
                 so   c  =  3 b/(2 + b²)         --- (2)

-b + a c + 2/ b = 0
       =>   b² + b c²  = 2                                ---(4)
       =>   b² - 2 + 9 b³ /(2 + b²)²  = 0
       =>   (b² - 2) (b² + 2)² + 9 b³ = 0          --- (3)
             
  Solving this:  we get:   b = 1     see directly from the equation.
                        so  c = 1       and   a = -1

factors are:  x² - x - 1          and   x² + x + 2

The factors for the first polynomial:   (x - (1+√5)/2)  (x - (1-√5)/2)
Roots  for the 2nd polynomial are imaginary....

so   x⁴ - 3 x - 2  = (x - (1+√5)/2) ( x - (1-√5)/2) (x² + x + 2)

Answer 2

P(x) = x⁴  - 3 x - 2

        = (x² + a x - b) ( x² + c x + 2 / b)             let  be....

        = x⁴ + (a+c) x³ + (-b+ a c +2/b) x²  + (-bc + 2a/b) x - 2 = 0

compare coefficients:

a = - c

b c - 2 a/b = 3   =>  b² c + 2 c - 3 b = 0          ---(1)

                 so   c  =  3 b/(2 + b²)         --- (2)

-b + a c + 2/ b = 0

       =>   b² + b c²  = 2                                ---(4)

       =>   b² - 2 + 9 b³ /(2 + b²)²  = 0

       =>   (b² - 2) (b² + 2)² + 9 b³ = 0          --- (3)

             

  Solving this:  we get:   b = 1     see directly from the equation.

                        so  c = 1       and   a = -1

factors are:  x² - x - 1          and   x² + x + 2

The factors for the first polynomial:   (x - (1+√5)/2)  (x - (1-√5)/2)

Roots  for the 2nd polynomial are imaginary....

so   x⁴ - 3 x - 2  = (x - (1+√5)/2) ( x - (1-√5)/2) (x² + x + 2)

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