Jeśli dwumiany x - 2 oraz x + 1 są dzielnikami wielomianu to zgodnie z twierdzeniem Bezouta:
f(2) = 0 oraz f(-1) = 0
[tex]\left \{ {{f(2)=0} \atop {f(-1)=0\\}} \right.\\\\\left \{ {{2*2^{3}+(a-3)*2^{2}+2a+b+6=0 } \atop {2*(-1)^{3}+(a-3)*(-1)^{2}+2a+b+6=0 }} \\\\\\[/tex]
[tex]\left \{ {{16+4(a-3)+2a+b+6=0} \atop {2+a-3+2a+b+6=0}} \right. \\\\\left \{ {{16+4a-12+2a+b+6=0} \atop {3a+b+5=0}} \right. \\\\\left \{ {{6a+b=-10} \atop {3a+b=-5}} \right. \\\\3a=-5/:3\\\\a=-\frac{5}{3} \\\\b=-5-3a=-5-3*(-\frac{5}{3} )=-5+5=0\\\\\left \{ {{a=-\frac{5}{3} } \atop {b=0}} \right.[/tex]
a = -5/3
b = 0
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Jeśli dwumiany x - 2 oraz x + 1 są dzielnikami wielomianu to zgodnie z twierdzeniem Bezouta:
f(2) = 0 oraz f(-1) = 0
[tex]\left \{ {{f(2)=0} \atop {f(-1)=0\\}} \right.\\\\\left \{ {{2*2^{3}+(a-3)*2^{2}+2a+b+6=0 } \atop {2*(-1)^{3}+(a-3)*(-1)^{2}+2a+b+6=0 }} \\\\\\[/tex]
[tex]\left \{ {{16+4(a-3)+2a+b+6=0} \atop {2+a-3+2a+b+6=0}} \right. \\\\\left \{ {{16+4a-12+2a+b+6=0} \atop {3a+b+5=0}} \right. \\\\\left \{ {{6a+b=-10} \atop {3a+b=-5}} \right. \\\\3a=-5/:3\\\\a=-\frac{5}{3} \\\\b=-5-3a=-5-3*(-\frac{5}{3} )=-5+5=0\\\\\left \{ {{a=-\frac{5}{3} } \atop {b=0}} \right.[/tex]
a = -5/3
b = 0