Question

Znajdź pierwiastki wielomianu W(x).

a)

b)

c)

Wskazówka. Przedstaw jeden z wyrazów wielomianu jako sumę dwóch jednomianów.

Answer 1

a)

W(x) = x⁴ - 4x³ + 8x² -24 x + 12 =

= x⁴ - 4x³ + 2x² + 6x² - 24x + 12 =

= x² (x² - 4x +2) + 6(x² - 4x +2 ) =

= (x² + 6)(x² -4x + 2)

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x² + 6 > 0 dla dowolnej liczby x ∈ R

x² - 4x + 2 = 0

Δ = (-4)² -4*1*2 = 16 - 8 = 8 = 4*2

√Δ = 2√2

x1 = [4 - 2√2]/2 = 2 - √2

x2 = [4 + 2√2]/2 = 2 + √2

Odp. 2 - √2 ; 2 + √2

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b)

W(x) = x⁴ - 3x³ + 5x² - 3x + 4 =

= x⁴ - 3 x³ + 4x² + x² - 3x + 4 =

= x²(x² -3x + 4) + (x² -3x + 4) =

= (x² + 1)( x² -3x +4) = 0 <=> x² - 3x + 4 = 0

bo

x² + 1 > 0  dla dowolnej liczby x ∈ R.

Δ = (-3)² - 4*1*4 = 9 - 16 = -7 < 0

Ten wielomian nie ma pierwiastków rzeczywistych.

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c)

W(x) = x⁴  + 3x³- x² -6x - 2 =

= x⁴ + 3x³ + x² - 2x² - 6x - 2 =

= x²(x²+3x + 1) - 2(x² + 3x + 1) =

= (x²-2)(x² +3x + 1) =

= (x + √2)(x - √2)(x² + 3x + 1) = 0 <=>

<=>   x +√2 = 0 ∨ x - √2 = 0 ∨ x²+3x+1 = 0 <=>

<=> x = - √2   ∨    x = √2    ∨   x = [-3 -√5]/2   ∨    x = [ -3 + √5]/2

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bo

Δ = 3²-4*1*1 = 9 - 4 = 5

√Δ = √5

x = [ -3 - √5]/2  ∨  x = [ -3 + √5]/2

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