Zad. 7
a)
[tex]w(x) = 4x^3 - \frac{1}{9} x= x(4x^2 - \frac{1}{9}) = x((2x)^2 - (\frac{1}{3})^2) =x(2x - \frac{1}{3})(2x + \frac{1}{3})[/tex]
Wzór: (a - b)(a + b) = a² - b²
b)
[tex]w(x) =16x^5-4x^3= 4x^3(4x^2 - 1) = 4x^3((2x)^2 - 1^2) = 4x^3(2x- 1)(2x+ 1)[/tex]
Wzór: (a - b)(a + b) = a² - b²
c)
[tex]w(x) = x^5-2x^4+5x^3= x^3(x^2 - 2x +5)[/tex]
x² - 2x + 5
a = 1, b = - 2, c = 5
Δ = (- 2)² - 4 · 1 · 5 = 4 - 20 = - 16 < 0, czyli trójmian kwadratowy x² - 2x + 5 nie ma postaci iloczynowej
d)
[tex]w(x) = 6x^3-15x^2 + 9x = 3x(2x^2- 5x+ 3) = 3x(x- 1)(2x - 3)[/tex]
2x² - 5x + 3
a = 2, b = - 5, c = 3
Δ = (- 5)² - 4 · 2 · 3 = 25 - 24 = 1; √Δ = √1 = 1
[tex]x_1 = \frac{-(-5)-1}{2 \cdot 2} =\frac{5-1}{4} =\frac{4}{4} = 1 \\ x_1 = \frac{-(-5)+1}{2 \cdot 2} =\frac{5+1}{4} =\frac{6}{4} =1,5[/tex]
2x² - 5x + 3 = 2(x - 1)(x - 1,5) = (x - 1)(2x - 3)
e)
[tex]x^4 - 5x^3 - 6x^2=x^2(x^2 - 5x - 6)=x^2(x+1)(x - 6)[/tex]
x² - 5x - 6
a = 1, b = - 5, c = - 6
Δ = (- 5)² - 4 · 1 · (- 6) = 25 + 24 = 49; √Δ = √49 = 7
[tex]x_1 = \frac{-(-5) -7}{2 \cdot 1} = \frac{5-7}{2} =\frac{-2}{2} = -1 \\ x_2 = \frac{-(-5)+7}{2 \cdot 1} = \frac{5+7}{2} =\frac{12}{2} =6[/tex]
x² - 5x - 6 = (x + 1)(x - 6)
f)
[tex]w(x) = (x^2 - 3x +2)(x^2 - 2x - 3)=(x-1)(x-2)(x+1)(x-3)[/tex]
x² - 3x + 2
a = 1, b = - 3, c = 2
Δ = (- 3)² - 4 · 1 · 2 = 9 - 8 = 1; √Δ = √1 = 1
[tex]x_1= \frac{-(-3) - 1}{2 \cdot 1} =\frac{3-1}{2} =\frac{2}{2} = 1 \\ x_2= \frac{-(-3) + 1}{2 \cdot 1} =\frac{3+1}{2} =\frac{4}{2} =2[/tex]
x² - 3x + 2 = (x - 1)(x - 2)
x² - 2x - 3
a = 1, b = - 2, c = - 3
Δ = (- 2)² - 4 · 1 · (- 3) = 4 + 12 = 16; √Δ = √16 = 4
[tex]x_1 = \frac{-(- 2) - 4}{2 \cdot 1} =\frac{2-4}2} =\frac{-2}{2} =-1 \\ x_2 = \frac{-(- 2) + 4}{2 \cdot 1} =\frac{2+4}2} =\frac{6}{2} =3[/tex]
x² - 2x - 3 = (x + 1)(x - 3)
g)
[tex]w(x) = x^4 + 2x^3 - 8x - 16 = x^3(x + 2) - 8(x + 2) = (x + 2)(x^3 - 8)= \\ = (x + 2)(x^3 - 2^3) = (x + 2)(x - 2)(x^2 + 2x + 4)[/tex]
Wzór: a³ - b³ = (a - b)(a² + ab + b²)
x² + 2x + 4
a = 1, b = 2, c = 4
Δ = 2² - 4 · 1 · 4 = 4 - 16 = - 12 < 0, czyli trójmian kwadratowy x² + 2x + 4 nie ma postaci iloczynowej
h)
[tex]w(x) = \frac{1}{2} x^3 - \frac{1}{6} x^2 - 3x + 1 = \frac{1}{2}x^2(x - \frac{1}{3}) - 3(x - \frac{1}{3}) =(x - \frac{1}{3})( \frac{1}{2}x^2 - 3) = \\ = \frac{1}{2}(x - \frac{1}{3})(x +\sqrt{6})(x - \sqrt{6)}[/tex]
¹/₂ x² - 3
a = ¹/₂, b = 0, c = - 3
Δ = 0² - 4 · ¹/₂ · (- 3) = 6; √Δ = √6
[tex]x_1 = \frac{0 - \sqrt{6}}{2 \cdot \frac{1}{2}} =\frac{-\sqrt{6}}{1} =-\sqrt{6} \\x_2 = \frac{0 +\sqrt{6}}{2 \cdot \frac{1}{2}} =\frac{\sqrt{6}}{1} =\sqrt{6}[/tex]
¹/₂ x² - 3 = ¹/₂(x + √6)(x - √6)
i)
[tex]w(x) = (20x^3 - 28x^2+ 8x)(x^4 +6x^3+2x^2+ 12) =4x(5x^2-7x+2) \cdot \\ (x^4 +6x^3+ 2x^2+ 12) =4x(5x-2)(x - 1)(x^4 +6x^3+ 2x^2+ 12)[/tex]
5x² - 7x + 2
a = 5, b = - 7, c = 2
Δ = (- 7)² - 4 · 5 · 2 = 49 - 40 = 9; √Δ = √9 = 3
[tex]x_1 = \frac{-(-7) - 3}{2 \cdot 5} =\frac{7-3}{10} =\frac{4}{10} =\frac{2}{5} \\ x_2= \frac{-(-7)+3}{2 \cdot 5} =\frac{7+3}{10} =\frac{10}{10} =1[/tex]
5x² - 7x + 2 = 5(x - ²/₅)(x - 1) = (5x - 2)(x - 1)
