Question

wyznacz pierwiastki wielomianu okreslonego wzorem

c) W(x)=(x²-4)(x+1)(x+5)

d) W(x)=(x²+3x)(x-2)(x+6)

Answer 1

c)

(x²-4)(x+1)(x+5)=0

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

x₁=2 x₂=-2 x₃=-1 x₄=-5

d)

(x²+3x)(x-2)(x+6)=0

x(x+3)(x-2)(x+6)=0

x₁=0 x₂=-3 x₃=2 x₄=-6

 

Answer 2

[x²-4][x+1][x+5]=0

x²=4

x₁=2

x₂=-2

x₃=-1

x₄=-5

 

[x²+3x][x-2][x+6]=0

x[x+3][x-2][x+6]=0

x₁=0

x₂=-3

x₃=2

x₄=-6