# How do you divide #(-x^5+7x^3-x)div(x^3-x^2+1)# using long division?

For the polynomial divison we can see it as;

Which will give us,

I will continue with the notation now,

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To divide (-x^5 + 7x^3 - x) by (x^3 - x^2 + 1) using long division:

- Divide the leading term of the dividend by the leading term of the divisor: (-x^5) ÷ (x^3) = -x^2.
- Multiply the entire divisor by the result from step 1: -x^2 * (x^3 - x^2 + 1) = -x^5 + x^4 - x^2.
- Subtract the result from step 2 from the dividend: (-x^5 + 7x^3 - x) - (-x^5 + x^4 - x^2) = 7x^3 - x^4 + x^2 - x.
- Repeat steps 1-3 with the new dividend: 7x^3 ÷ x^3 = 7.
- Multiply the entire divisor by the result from step 4: 7 * (x^3 - x^2 + 1) = 7x^3 - 7x^2 + 7.
- Subtract the result from step 5 from the dividend: (7x^3 - x^4 + x^2 - x) - (7x^3 - 7x^2 + 7) = -x^4 + 8x^2 - x - 7.
- Repeat steps 1-3 with the new dividend: -x^4 ÷ x^3 = -x.
- Multiply the entire divisor by the result from step 7: -x * (x^3 - x^2 + 1) = -x^4 + x^3 - x.
- Subtract the result from step 8 from the dividend: (-x^4 + 8x^2 - x - 7) - (-x^4 + x^3 - x) = 8x^2 - x^3 - 7.
- Repeat steps 1-3 with the new dividend: 8x^2 ÷ x^3 = 0 (as the degree of the dividend is less than the degree of the divisor).

The division process stops here. Therefore, the quotient is -x^2 + 7 - x, and the remainder is 8x^2 - x^3 - 7.

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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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