(4x^4-4x^2-x-3)divided by(2x^2-3)
quotient an remainder?
To find the quotient and remainder of the division of polynomials, we can use long division.
Step 1: First, let's set up the long division:
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2x^2 - 3 | 4x^4 - 4x^2 - x - 3
Step 2: Divide the first term of the dividend (4x^4) by the first term of the divisor (2x^2) to get 2x^2.
2x^2
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2x^2 - 3 | 4x^4 - 4x^2 - x - 3
Step 3: Multiply the entire divisor (2x^2 - 3) by 2x^2:
2x^2
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2x^2 - 3 | 4x^4 - 4x^2 - x - 3
- (4x^4 - 6x^2)
Step 4: Subtract the result from the previous step from the dividend:
2x^2
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2x^2 - 3 | 4x^4 - 4x^2 - x - 3
- (4x^4 - 6x^2)
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2x^2 - x - 3
Step 5: Bring down the next term of the dividend, which is -x:
2x^2 + (-1x)
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2x^2 - 3 | 4x^4 - 4x^2 - x - 3
- (4x^4 - 6x^2)
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2x^2 - x - 3
- (2x^2 - x)
Step 6: Repeat steps 2-5 with the new polynomial 2x^2 - x - 3:
2x^2 + (-1x)
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2x^2 - 3 | 4x^4 - 4x^2 - x - 3
- (4x^4 - 6x^2)
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2x^2 - x - 3
- (2x^2 - x)
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0
Step 7: Since the remainder is zero, we stop the long division process here.
Therefore, the quotient is 2x^2 + (-1x) and the remainder is 0.