# How do you divide #(x^3+x+3)/(x-5)#?

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The remainder is

Let's perform a synthetic division

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Quotient

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To divide (x^3+x+3) by (x-5), you can use long division or synthetic division. Here is the solution using long division:

- Divide x^3 by x, which gives x^2.
- Multiply (x-5) by x^2, which gives x^3-5x^2.
- Subtract (x^3-5x^2) from (x^3+x+3), which gives 6x^2+x+3.
- Divide 6x^2 by x, which gives 6x.
- Multiply (x-5) by 6x, which gives 6x^2-30x.
- Subtract (6x^2-30x) from (6x^2+x+3), which gives 31x+3.
- Divide 31x by x, which gives 31.
- Multiply (x-5) by 31, which gives 31x-155.
- Subtract (31x-155) from (31x+3), which gives 158.
- Since the degree of 158 is less than the degree of (x-5), the division is complete.

Therefore, the quotient is x^2+6x+31 and the remainder is 158.

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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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