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What are the roots of the polynomial equation x4+x3=4x2+4x? Use a graphing calculator and a system of equations.

A.)–2, –1, 0, 2
B.)–2, 0, 1, 2
C.)–1, 0
D.)0, 1

Respuesta :

WojtekR
[tex]x^4+x^3=4x^2+4x\\\\x^4+x^3-4x^2-4x=0\\\\x\Big[x^3+x^2-4x-4\Big]=0\\\\ x\Big[x^2\big(x+1\big)-4\big(x+1\big)\Big]=0\\\\x\Big[\big(x+1\big)\big(x^2-4\big)\Big]=0\\\\x\big(x+1\big)\big(x^2-4\big)=0\\\\ x\big(x+1\big)\big(x+2\big)\big(x-2\big)=0\\\\\\x=0\quad\vee\quad x+1=0\quad\vee\quad x+2=0\quad\vee\quad x-2=0\\\\ \boxed{x=0\quad\vee\quad x=-1\quad\vee\quad x=-2\quad\vee\quad x=2}[/tex]

Answer A.

Answer:

A. –2, –1, 0, 2

Step-by-step explanation:

Given : [tex]x^{4} +x^{3} = 4x^{2} +4x[/tex]

Solution:

[tex]x^{4} +x^{3} -4x^{2} -4x=0[/tex]


[tex]x(x^{3} +x^{2} -4x-4)[/tex] (taking x common from the equation)


[tex]x[x^{2} (x+1) - 4(x+1)]=0[/tex] (taking x+1 common in thye square bracket )


then, [tex]x[(x+1)(x^{2}-4)] =0[/tex]


then , [tex](x)(x+1)(x^{2}-4) =0[/tex]


[tex](x)(x+1)(x+2)(x-2)[/tex]


⇒[tex]x = 0[/tex] ,   [tex]x+1 =0[/tex] ,   [tex]x+2 =0[/tex] ,   [tex]x-2=0[/tex]


⇒ [tex]x=0[/tex] ,   [tex]x= -1 [/tex] ,   [tex]x = -2[/tex] ,   [tex]x=2[/tex]

So, the roots of the polynomial equation [tex]x^{4} +x^{3} = 4x^{2} +4x[/tex] are –2, –1, 0, 2 (option A)

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