Question

Proszę o pomoc, nikt nie wie jak to rozwalić. Muszę rozłożyć te wielomiany na czynniki.

A

B

G

F

Answer 1

W(x) = x³-4x+3

W(x) = x³-x-3x+3

W(x)= x(x²-1)-3(x-1)

W(x)=x(x-1)(x+1)-3(x-1)

W(x)= (x-1)[x(x+1)-3]

W(x) = (x-1)(x²+x-3)

 

Δ=1-4*(-3) = 1+12=13

 

 

 

 

B)

W(x) = x³  +2x² -3

W(x)= x³-1+2x² - 2

W(x) = (x-1)(x²+x+1)+2(x²-1)

W(x) = (x-1)(x²+x+1)+2(x-1)(x+1)

W(x) = (x-1)[x²+x+1+2(x+1)]

W(x) = (x-1)(x²+x+1+2x+2)

W(x) = (x-1)(x²+3x+3)

Δ=9-12<0

więc zostawiamy tak jak jest, bo nie da się już 2 nawiasu rozpisać

 

G)

W(x) = 2x⁴ -3x³ -2x² + 3x

W(x) = 2x²(x²-1)-3x(x²-1)

W(x)=(2x²-3x)(x²-1)

W(x) = 2x(x-3)(x²-1)

W(x) = 2x(x-3)(x-1)(x+1)

 

F)

W(x) = x⁵ - 3x⁴ -x³ +3x² -6x +18

W(x) = x³(x²-1)-3x²(x²-1)-6(x-3)

W(x) = (x²-1)(x³-3x²)-6(x-3)

W(x)=(x-1)(x+1)*x²(x-3)-6(x-3)

W(x)=(x-3)[(x-1)(x+1)x²-6]

W(x) = (x-3)[(x²-1)x²-6]

W(x)=(x-3)(x⁴-x²-6)

teraz ropisujemy x⁴-x²-6

niech x²=t

t²-t-6=0

Δ=1+24=25

t₁=(1+5)/2=3

t₂=(1-5)/2=-2

 

(t²-t-6)=(t-3)(t+2)

x²=t

(x²-3)(x²+2)=(x+√3)(x-√3)(x²+2)

 

W(x)=(x-3)(x-√3)(x+√3)(x²+2)