how to tolve the long division for (x^5-5x-3x^3+4x^2+2x^4+2)÷(1-X)

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Asked 7/8/2009 2:02:36 PM

Updated 8/11/2023 2:18:34 PM

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Expert answered|AndrewD|Points 171|

Question

Asked 7/8/2009 2:02:36 PM

Updated 8/11/2023 2:18:34 PM

1 Answer/Comment

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1. Divide the first term of the dividend (x^5) by the divisor (1) to get x^5.

2. Multiply the divisor (1) by x^5 to get x^5.

3. Subtract x^5 from the dividend (x^5 - 5x^5) to get -5x^5.

4. Bring down the next term, which is -3x^3, to the remainder.

5. Divide -5x^5 by the divisor (1) to get -5x^5.

6. Multiply the divisor (1) by -5x^5 to get -5x^5.

7. Subtract -5x^5 from the remainder (-5x^5 - (-5x^5)) to get 0.

8. Since the remainder is now 0, you've finished the division.

The result of the division is x^5 - 5x^5 - 3x^3 + 4x^2 + 2x^4 + 2, with no remainder.

So, (x^5 - 5x - 3x^3 + 4x^2 + 2x^4 + 2) ÷ (1 - x) = x^5 - 5x^5 - 3x^3 + 4x^2 + 2x^4 + 2.

2. Multiply the divisor (1) by x^5 to get x^5.

3. Subtract x^5 from the dividend (x^5 - 5x^5) to get -5x^5.

4. Bring down the next term, which is -3x^3, to the remainder.

5. Divide -5x^5 by the divisor (1) to get -5x^5.

6. Multiply the divisor (1) by -5x^5 to get -5x^5.

7. Subtract -5x^5 from the remainder (-5x^5 - (-5x^5)) to get 0.

8. Since the remainder is now 0, you've finished the division.

The result of the division is x^5 - 5x^5 - 3x^3 + 4x^2 + 2x^4 + 2, with no remainder.

So, (x^5 - 5x - 3x^3 + 4x^2 + 2x^4 + 2) ÷ (1 - x) = x^5 - 5x^5 - 3x^3 + 4x^2 + 2x^4 + 2.

Added 8/11/2023 2:18:28 PM

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