To find the quotient of the polynomial \((5x^4 - 10x^3 - 13x^2 + 12x + 30)\) divided by \((x^2 - 4x + 4)\), we can use polynomial long division.
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Set up the division: We want to divide \(5x^4 - 10x^3 - 13x^2 + 12x + 30\) by \(x^2 - 4x + 4\).
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First Division: Divide the leading term of the dividend \(5x^4\) by the leading term of the divisor \(x^2\): \[ \frac{5x^4}{x^2} = 5x^2 \] Multiply \(5x^2\) by the entire divisor \(x^2 - 4x + 4\): \[ 5x^2(x^2 - 4x + 4) = 5x^4 - 20x^3 + 20x^2 \] Subtract this result from the original polynomial: \[ (5x^4 - 10x^3 - 13x^2 + 12x + 30) - (5x^4 - 20x^3 + 20x^2) = (10x^3 - 33x^2 + 12x + 30) \]
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Second Division: Now divide the leading term \(10x^3\) by \(x^2\): \[ \frac{10x^3}{x^2} = 10x \] Multiply \(10x\) by the entire divisor \(x^2 - 4x + 4\): \[ 10x(x^2 - 4x + 4) = 10x^3 - 40x^2 + 40x \] Now, subtract this from the current polynomial: \[ (10x^3 - 33x^2 + 12x + 30) - (10x^3 - 40x^2 + 40x) = (7x^2 - 28x + 30) \]
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Third Division: Divide the leading term \(7x^2\) by \(x^2\): \[ \frac{7x^2}{x^2} = 7 \] Multiply \(7\) by the entire divisor \(x^2 - 4x + 4\): \[ 7(x^2 - 4x + 4) = 7x^2 - 28x + 28 \] Subtract this from the current polynomial: \[ (7x^2 - 28x + 30) - (7x^2 - 28x + 28) = 2 \]
At this point, we can summarize our results. The quotient of the division is: \[ 5x^2 + 10x + 7 \] and the remainder is \(2\).
Thus, the final answer is: \[ \text{Quotient: } 5x^2 + 10x + 7 \quad \text{Remainder: } 2 \]