Oblicz wartość wielomianu W(x) dla x=x0

a) W(x)=x^4-2x^2+√3x-7 , x0=√3

b) w(x)=x^6-√2x^5+x^4-2√2x^3+x+4 , x0=√2

Proszę o pilna pomoc w obliczeniu



Odpowiedź :

ZbiorJ

[tex]a)~~W(x) =x^{4} -2x^{2} +\sqrt{3} \cdot x -7~~\land~~ x_{0} =\sqrt{3} \\\\W(\sqrt{3} ) =(\sqrt{3} )^{4} -2(\sqrt{3} )^{2} +\sqrt{3} \cdot \sqrt{3} -7\\\\W(\sqrt{3} )=3^{\frac{1}{2} \cdot 4} -2 \cdot 3^{\frac{1}{2} \cdot 2}+3-7\\\\W(\sqrt{3} )=3^{2} -2\cdot 3+3-7\\\\W(\sqrt{3} )=9-6+3-7\\\\W(\sqrt{3} )= -1[/tex]

[tex]b)~~W(x)=x^{6} -\sqrt{2} \cdot x^{5} +x^{4} -2\sqrt{2} \cdot x^{3} +x+4~~\land ~~x_{0} =\sqrt{2} \\\\W(\sqrt{2})=(\sqrt{2}) ^{6} -\sqrt{2} \cdot ( \sqrt{2})^{5} +( \sqrt{2})^{4} -2\sqrt{2} \cdot ( \sqrt{2})^{3} +\sqrt{2} +4\\\\W(\sqrt{2})=(\sqrt{2}) ^{6} -(\sqrt{2}) ^{1+5} +2^{\frac{1}{2} \cdot 4 } - 2\cdot (\sqrt{2}) ^{1+3} +\sqrt{2} +4\\\\W(\sqrt{2})=(\sqrt{2}) ^{6} -(\sqrt{2}) ^{6} +2^{2} -2\cdot 2^{\frac{1}{2} \cdot 4 } + \sqrt{2} +4\\\\W(\sqrt{2} )=4-2\cdot 2^{2} +\sqrt{2} +4[/tex]

[tex]W(\sqrt{2} )=4-2\cdot 4 +\sqrt{2} +4\\\\W(\sqrt{2} )=4-8 +\sqrt{2} +4\\\\W(\sqrt{2} )=-4+\sqrt{2} +4\\\\W(\sqrt{2} )=\sqrt{2}[/tex]