# How do you divide #(2x^3+7x^2-5x-4) / (2x+1)# using polynomial long division?

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To divide (2x^3+7x^2-5x-4) by (2x+1) using polynomial long division, follow these steps:

- Arrange the dividend (2x^3+7x^2-5x-4) and the divisor (2x+1) in descending order of exponents.
- Divide the first term of the dividend (2x^3) by the first term of the divisor (2x). The result is x^2.
- Multiply the divisor (2x+1) by the quotient obtained in step 2 (x^2). The result is 2x^3+x^2.
- Subtract the product obtained in step 3 (2x^3+x^2) from the dividend (2x^3+7x^2-5x-4). The result is 6x^2-5x-4.
- Bring down the next term from the dividend (-5x).
- Divide the first term of the new dividend (6x^2) by the first term of the divisor (2x). The result is 3x.
- Multiply the divisor (2x+1) by the quotient obtained in step 6 (3x). The result is 6x^2+3x.
- Subtract the product obtained in step 7 (6x^2+3x) from the new dividend (6x^2-5x-4). The result is -8x-4.
- Bring down the next term from the dividend (-4).
- Divide the first term of the new dividend (-8x) by the first term of the divisor (2x). The result is -4.
- Multiply the divisor (2x+1) by the quotient obtained in step 10 (-4). The result is -8x-4.
- Subtract the product obtained in step 11 (-8x-4) from the new dividend (-8x-4). The result is 0.

The quotient is x^2 + 3x - 4, and the remainder is 0.

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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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