Linearfaktorzerlegung von Polynomen. Bsp.5. y= x^2/3 + 2x

Question

2 Antworten

Ein anderes Problem?

Grosserloewe · Answer 1 · 2017-10-26T19:15:37+0000

zu 2.)

=(2/3) *x^4 -(22/3) x^3 -(48/3)*x^2

Klammere 2/3 x^2 aus

------>

=2/3 x^2 (x^2 -11x +24)

----->x^2 -11x +24 =0 z.B. pq - Formel

x_1.2= 11/2±√ (121/2 -96/4)

x_1.2= 11/2±5/2

x₁= 8

x₂=3

--------->

=(2/3) x^2 (x-8)( x-3)

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

5.)

x^2/3 +2x

=x( (x/3) +2)

Gast jc2144 · Answer 2 · 2017-10-27T08:46:01+0000

1) 2x^4/3+6x^3=

x^3(2x/3 +6)=2/3*x^3(x+9)