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Answer:
[tex]x=-1,\:x=-7,\:x=i,\:x=-i[/tex]
Step-by-step explanation:
Considering the equation
[tex]x^4+8x^3+8x^2+8x+7=0[/tex]
Solving
[tex]x^4+8x^3+8x^2+8x+7[/tex]
[tex]\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)[/tex]
As
[tex]\mathrm{Use\:the\:rational\:root\:theorem}[/tex]
[tex]a_0=7,\:\quad a_n=1[/tex]
[tex]\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1[/tex]
[tex]\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}[/tex]
[tex]-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1[/tex]
[tex]=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A][/tex]
Solving
[tex]\frac{x^4+8x^3+8x^2+8x+7}{x+1}[/tex]
[tex]=x^3+7x^2+x+7[/tex]
Putting [tex]\frac{x^4+8x^3+8x^2+8x+7}{x+1}[/tex] = [tex]x^3+7x^2+x+7[/tex] in equation [A]
So,
[tex]\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A][/tex]
[tex]=\left(x+1\right)x^3+7x^2+x+7[/tex]
As
[tex]x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)[/tex]
So,
Equation [A] becomes
[tex]=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)[/tex]
So, the polynomial equation becomes
[tex]\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0[/tex]
[tex]\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex][tex]\mathrm{Solve\:}\:x+1=0:\quad x=-1[/tex]
[tex]\mathrm{Solve\:}\:x+7=0:\quad x=-7[/tex]
[tex]\mathrm{Solve\:}\:x^2+1=0:\quad x=i,\:x=-i[/tex]
[tex]\mathrm{The\:solutions\:are}[/tex]
[tex]x=-1,\:x=-7,\:x=i,\:x=-i[/tex]
Keywords: polynomial equation
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