Polynomdivision anwenden: f(x)=0,25x^4-0,25x^3-2x^2+3x

Question

Brauch Hilfe,

 

Also bestimmen Sie die Achsenschnittpunkte sowie die Linearfaktordarstellung mithilfe der Polynomdivision.

 

a) f(x)=0,25x⁴-0,25x³-2x²+3x

b) f(x)= 0,25x⁴-x³+4x+2,75

 

Answer 1

Hi,

a) f(x)=0,25x⁴-0,25x³-2x²+3x

= x(0,25x^3-0,25x^2-2x+3)

Raten einer weiteren Nullstelle (die erste ist x₁ = 0): x₂ = 2

(1/4x^3  - 1/4x^2  -   2x  + 3) : (x - 2)  =  1/4x^2 + 1/4x - 3/2 

-(1/4x^3  - 1/2x^2)
----------------------------
           1/4x^2  -   2x  + 3
         -(1/4x^2  - 1/2x)    
      --------------------------
                   - 3/2x  + 3
                 -(- 3/2x  + 3)
                   ------------------
                                  0

 

1/4x^2 + 1/4x - 3/2 = 0    |*4, dann pq-Formel

x₃ = -3 und x₄ = 2

 

-->

f(x) = x(0,25x^3-0,25x^2-2x+3) = 1/4*x(x+3)(x-2)^2

Die Nullstellen sind bereits genannt. Der y-Achsenschnittpunkt ist S(0|0).

 

b) f(x)= 0,25x⁴-x³+4x+2,75

Raten von x₁ = -1

(1/4x^4  -    x^3            +    4x  + 11/4) : (x + 1)  =  1/4x^3 - 5/4x^2 + 5/4x + 11/4  
-(1/4x^4  + 1/4x^3)                          
------------------------------------------------
         - 5/4x^3            +    4x  + 11/4
       -(- 5/4x^3  - 5/4x^2)                
         -----------------------------------------
                     5/4x^2  +    4x  + 11/4
                   -(5/4x^2  +  5/4x)       
                     --------------------------------------
                               11/4x  + 11/4
                             -(11/4x  + 11/4)
                               -----------------------
                                           0

 

Erneutes Raten von x₂ = -1

(1/4x^3  - 5/4x^2  +  5/4x  + 11/4) : (x + 1)  =  1/4x^2 - 3/2x + 11/4  
-(1/4x^3  + 1/4x^2)                
----------------------------------------
         - 3/2x^2  +  5/4x  + 11/4
       -(- 3/2x^2  -  3/2x)       
         -------------------------------------
                     11/4x  + 11/4
                   -(11/4x  + 11/4)
------------------------------------ 
                                      0

 

1/4x^2 - 3/2x + 11/4  = 0   |*4, pq-Formel

-> keine weitere Lösungen.

Faktorezerlegung:

f(x)= 0,25x⁴-x³+4x+2,75 = 0,25(x+1)^2(x^2-6x+11)

Nullstellen sind x_1,2 = -1

y-Achsenschnittpunkt ist S(0|2,75)

 

Grüße