Question

Rozwiąż równania wielomianowe:

(coś nie wyszło z wpisywaniem- liczby po X to potęgi jakby co)

1) X4 - 1 = 0

2) X6- 64 = 0

3) (x3 – 5)2 – 36 = 0

4) X3 + 3x – 4 = 0

5) X3 + 4x2 – 2x – 8 = 0

6) X3 – 3x2 + 4x – 12 = 0

7) 2x5 + 3x4 – 2x – 3 = 0

8) 5x5 + 4x4 – 5x – 4 = 0

9) (2x – 5)(9x2 – 4) = (9x2 – 4)(x – 8)

Answer 1

1) x^4 - 1 = 0

(x^2 - 1)(x^2 + 1) = 0

(x-1)(x+1)(x^2 + 1) = 0

x=1 v x=-1

 

2) x^6 - 64 = 0

[(x^3)^2 - 8^2] = 0

(x^3 - 8)(x^3 + 8)=0

(x-2)(x^2 + x+4)(x+2)(x^2 - x+4)=0

x=2 v x=-2

 

3) (x^3 - 5)^2 - 36 = 0

(x^3 - 5)^2 - 6^2 = 0

(x^3 - 5 - 6)(x^3 - 5 + 6) = 0

(x^3 - 11)(x^3 + 1) = 0

(x-pierw.3.st.z.11)(x^2+<pierw.3.stopnia.z.11>x + pierw.3.st.z.121)(x+1)(x^2-x+1)=0

x=pierw.3.st.z.11 v x=-1

 

4) x^3 + 3x - 4 = 0

x^3 - x + 4x - 4 = 0

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

x(x-1)(x+1) + 4(x-1) = 0

(x-1) [ x(x+1) + 4 ] = 0

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

x=1

 

5) x^2 + 4x^2 - 2x - 8 = 0

x^2(x+4) - 2(x+4) = 0

(x^2 - 2)(x+4) = 0

(x-pierw.z.2)(x+pierw.z.2)(x+4)=0

x=pierw.z.2 v x=-pierw.z.2 v x=-4

 

6) x^3 - 3x^2 + 4x - 12 =0

x^2(x-3)-4(x-3)=0

(x^2 - 4)(x-3)=0

(x-2)(x+2)(x-3)=0

x=2 v x=-2 v x=3

 

7) 2x^5 + 3x^4 - 2x - 3 = 0

x^4(2x+3) - (2x+3) = 0

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

(x^2 - 1)(x^2 + 1)(2x+3) = 0

(x-1)(x+1)(x^2 + 1)(2x+3)=0

x=1 v x=-1 v x=-1,5

 

8) 5x^5 + 4x^4 - 5x - 4 = 0

x^4(5x+4) - (5x+4) = 0

(x^4 - 1)(5x+4) = 0

(x^2 - 1)(x^2 + 1)(5x+4)=0

(x-1)(x+1)(x^2 + 1)(5x+4)=0

x=1 v x=-1 v x=-4/5

 

9) (2x-5)(9x^2-4) = (9x^2-4)(x-8)

(2x-5)(3x-2)(3x+2) - (3x-2)(3x+2)(x-8) = 0

(3x-2)(3x+2) [(2x-5)-(x-8)]=0

(3x-2)(3x+2)(2x-5-x+8)=0

(3x-2)(3x+2)(x+3)=0

x=2/3 v x=-2/3 v x=-3