# How do you factor #x^4+x^2+1#?

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This really makes a bit more sense in the complex numbers...

First note that:

graph{((x-1)^2+(y-0)^2-0.002)((x-1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1)^2+(y-0)^2-0.002)((x+1/2)^2+(y+sqrt(3)/2)^2-0.002)((x-1/2)^2+(y+sqrt(3)/2)^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}

They form the vertices of a regular hexagon.

de Moivre's formula tells us that:

For instance we find:

Hence:

For example:

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The difference of squares identity can be written:

The difference of cubes identity can be written:

The sum of cubes identity can be written:

Note that:

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To factor the expression x^4 + x^2 + 1, you can treat it as a quadratic in terms of x^2. Factoring it as such, you get: (x^2 + 1)(x^2 + 1). Therefore, the factored form is (x^2 + 1)^2.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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