If y=x+1x, then x^4+x^3-4x^2+x+1=0 becomes:

1953 AMC 12 Problem 20

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20.

y=x+\dfrac1x,

then

x^4+x^3-4x^2+x+1=0

becomes:

x^2(y^2+y-2)=0

x^2(y^2+y-3)=0

x^2(y^2+y-4)=0

x^2(y^2+y-6)=0

none of these

Factor out

x^2

and group reciprocal terms

Use

x^2+x^{-2}=(x+x^{-1})^2-2

Because

x\ne0,

factor the expression as

\begin{aligned} x^2\Bigg[& \left(x^2+\frac1{x^2}\right)\\ &+\left(x+\frac1x\right)-4 \Bigg]. \end{aligned}

Since

x^2+x^{-2}=y^2-2,

this becomes

x^2(y^2+y-6)=0.

Thus, the correct answer is D.