Odpowiedź :
Zadanie 3.
Sprytny sposób na policzenie sumy współczynników wielomianu to policzenie wartości wielomianu dla x=1.
a)
[tex]u(1)=(2*1^2-1)^2-(2*1-1)^2=1^2-1^2=0[/tex]
b)
[tex]u(1)=(1^2+8)(\sqrt8-1)(1+2\sqrt2)=(1+8)(2\sqrt2-1)(1+2\sqrt2)=9*(8-1)=63[/tex]
Uwaga: Tu użyłem wzoru [tex](a-b)(a+b)=a^2-b^2[/tex].
c)
[tex]u(1)=(1-2)^3-(1-2)^2=(-1)^3-(-1)^2=-1-1=-2[/tex]
d)
[tex]u(1)=(1^2+1+1)(1^2-1+1)=3*1=3[/tex]
Zadanie 4.
a)
[tex]w(x)=3x^4-3x^2=3x^2(x^2-1)=3x^2(x-1)(x+1)[/tex]
b)
[tex]w(x)=2x^3+4x^2+2x=2x(x^2+2x+1)=2x(x+1)^2[/tex]
c)
[tex]w(x)=x^6+7x^5+6x^4=x^4(x^2+7x+6)\\\Delta=7^2-4*1*6=49-24=25\\\sqrt\Delta=5\\x_1=\frac{-7-5}{2}=-6\\x_2=\frac{-7+5}{2}=-1\\w(x)=x^4(x+6)(x+1)[/tex]
d)
[tex]w(x)=5x^5-10x^3+5x=5x(x^4-2x^2+1)=5x(x^2-1)^2=5x(x-1)^2(x+1)^2[/tex]
e)
[tex]w(x)=-3x^5+30x^3-75x=-3x(x^4-10x^2+25)=-3x(x^2-5)^2=\\=-3x(x-\sqrt5)^2(x+\sqrt5)^2[/tex]
f)
[tex]w(x)=32x^6-16x^4+2x^2=2x^2(16x^4-8x^2+1)=2x^2(4x^2-1)^2=\\=2x^2(2x-1)^2(2x+1)^2[/tex]
Zadanie 5.
[tex]w(-1)=4\qquad w(0)=3[/tex]
a)
[tex]\left \{ {{(-1)^3+(a-b)*(-1)^2-4*(-1)+\frac{b}{2}=4} \atop {0^3+(a-b)*0^2-4*0+\frac{b}{2}=3}} \right. \\\left \{ {{-1+a-b+4+\frac{b}{2}=4} \atop {\frac{b}{2}=3\ |*2}} \right. \\\left \{ {{a-b+\frac{b}{2}=1} \atop b=6}} \right. \\\left \{ {{a-6+\frac{6}{2}=1} \atop b=6}} \right. \\\left \{ {{a-6+3=1} \atop b=6}} \right. \\\left \{ {{a=4} \atop b=6}} \right.[/tex]
b)
[tex]\left \{ {{-a*(-1)^3+3*(-1)^2+a-6b=4} \atop {-4*0^3+3*0^2+a-6b=3}} \right. \\\left \{ {{a+3+a-6b=4} \atop {a-6b=3}} \right. \\\left \{ {{2a-6b=1} \atop {a=6b+3}} \right. \\\left \{ {{12b+6-6b=1} \atop {a=6b+3}} \right. \\\left \{ {{6b=-5\ |:6} \atop {a=6b+3}} \right. \\\left \{ {{b=-\frac{5}{6}} \atop {a=6*(-\frac{5}{6})+3}} \right. \\\left \{ {{b=-\frac{5}{6}} \atop {a=-2}} \right.[/tex]
Zadanie 6.
a)
[tex]5x^2+2x^4+x^3=0\\2x^4+x^3+5x^2=0\\x^2(2x^2+x+5)=0\\x^2=0\ \vee\ 2x^2+x+5=0\\x=0\ \vee\ 2x^2+x+5=0\\\Delta=1^2-4*2*5=1-40=-39 < 0\\\text{Drugie r\'ownanie jest sprzeczne. Ostatecznie}\\x=0[/tex]
b)
[tex]-2x^3-6x^2+8x=0\ |:(-2)\\x^3+3x^2-4x=0\\x(x^2+3x-4)=0\\x=0\ \vee\ x^2+3x-4=0\\\Delta=3^2-4*1*(-4)=9+16=25\\\sqrt\Delta=5\\x_1=\frac{-3-5}{2}=-4\\x_2=\frac{-3+5}{2}=1\\\text{Ostatecznie}\\x\in\{-4,0,1\}[/tex]
c)
[tex]x^5-7x^4+12x^3=0\\x^3(x^2-7x+12)=0\\x^3=0\ \vee\ x^2-7x+12=0\\x=0\ \vee\ x^2-7x+12=0\\\Delta=(-7)^2-4*1*12=49-48=1\\\sqrt\Delta=1\\x_1=\frac{7-1}{2}=3\\x_2=\frac{7+1}{2}=4\\\text{Ostatecznie}\\x\in\{0,3,4\}[/tex]
d)
[tex]20x^6+x^5=x^4\\20x^6+x^5-x^4=0\\x^4(20x^2+x-1)=0\\x^4=0\ \vee\ 20x^2+x-1=0\\x=0\ \vee\ 20x^2+x-1=0\\\Delta=1^1-4*20*(-1)=1+80=81\\\sqrt\Delta=9\\x_1=\frac{-1-9}{2*20}=\frac{-10}{40}=-\frac{1}{4}\\x_2=\frac{-1+9}{2*20}=\frac{8}{40}=\frac{1}{5}\\\text{Ostatecznie}\\x\in\{-\frac{1}{4},0,\frac{1}{5}\}[/tex]
e)
[tex]4x^5+x^3=4x^4\\4x^5-4x^4+x^3=0\\x^3(4x^2-4x+1)=0\\x^3(2x-1)^2=0\\x^3=0\ \vee\ (2x-1)^2=0\\x=0\ \vee\ 2x-1=0\\x=0\ \vee\ 2x=1\ |:2\\x=0\ \vee\ x=\frac{1}{2}\\\text{Ostatecznie}\\x\in\{0,\frac{1}{2}\}[/tex]
f)
[tex]x^4=2x^6+x^5\\2x^6+x^5-x^4=0\\x^4(2x^2+x-1)=0\\x^4=0\ \vee\ 2x^2+x-1=0\\x=0\ \vee\ 2x^2+x-1=0\\\Delta=1^2-4*2*(-1)=1+8=9\\\sqrt\Delta=3\\x_1=\frac{-1-3}{2*2}=\frac{-4}{4}=-1\\x_2=\frac{-1+3}{2*2}=\frac{2}{4}=\frac{1}{2}\\\text{Ostatecznie}\\x\in\{-1,0,\frac{1}{2}\}[/tex]
Zadanie 8.
a)
[tex]w(x)=x^3-2x^2-2x+4=x^2(x-2)-2(x-2)=(x-2)(x^2-2)=\\=(x-2)(x-\sqrt2)(x+\sqrt2)\\\text{Pierwiastki}: -\sqrt2,\ \sqrt2,\ 2[/tex]
b)
[tex]w(x)=x^4+x^3-4x^2-4x=x(x^3+x^2-4x-4)=x[x^2(x+1)-4(x+1)]=\\=x(x+1)(x^2-4)=x(x+1)(x-2)(x+2)\\\text{Pierwiastki}:\ -2,-1,0,2[/tex]
c)
[tex]w(x)=5x^3+x^2-15x-3=x^2(5x+1)-3(5x+1)=(5x+1)(x^2-3)=\\=(5x+1)(x-\sqrt3)(x+\sqrt3)\\\text{Pierwiastki}:\ -\sqrt3,-\frac{1}{5},\sqrt3[/tex]
d)
[tex]w(x)=125x^3-27=(5x-3)(25x^2+15x+9)\\\text{Pierwiastki}:\ \frac{3}{5}[/tex]
e)
[tex]w(x)=27x^4+\frac{1}{8}x=\frac{1}{8}x(216x^3+1)=\frac{1}{8}x(6x+1)(36x^2-6x+1)\\\text{Pierwiastki}:-\frac{1}{6},0[/tex]
f)
[tex]w(x)=-14x^3+7x=-7x(2x^2-1)=-7x(\sqrt2x-1)(\sqrt2x+1)\\\text{Pierwiastki}:\ -\frac{\sqrt2}{2},0,\frac{\sqrt2}{2}[/tex]