QUESTION IMAGE
Question
- $(3x^2 - 3x + 5) div (x - 6)$ 7. $(4x^3 - 6x^2 - 4x + 8) div (2x - 1)$ 6. $(2x^3 + 6x^2 - 6x + 2) div (2x - 3)$ 8. $(2x^3 - 5x^2 + 3x - 6) div (x + 4)$
Problem 5: \((3x^2 - 3x + 5) \div (x - 6)\)
Step 1: Use polynomial long division
Divide the first term of the dividend \(3x^2\) by the first term of the divisor \(x\) to get \(3x\). Multiply the divisor \((x - 6)\) by \(3x\) to get \(3x^2 - 18x\). Subtract this from the dividend:
Step 2: Divide the new first term \(15x\) by \(x\) to get \(15\). Multiply the divisor \((x - 6)\) by \(15\) to get \(15x - 90\). Subtract this from \(15x + 5\):
So, \((3x^2 - 3x + 5) \div (x - 6) = 3x + 15 + \frac{95}{x - 6}\)
Problem 6: \((2x^3 + 6x^2 - 6x + 2) \div (2x - 3)\)
Step 1: Divide the first term of the dividend \(2x^3\) by the first term of the divisor \(2x\) to get \(x^2\). Multiply the divisor \((2x - 3)\) by \(x^2\) to get \(2x^3 - 3x^2\). Subtract this from the dividend:
Step 2: Divide the new first term \(9x^2\) by \(2x\) to get \(\frac{9}{2}x\). Multiply the divisor \((2x - 3)\) by \(\frac{9}{2}x\) to get \(9x^2 - \frac{27}{2}x\). Subtract this from \(9x^2 - 6x + 2\):
Step 3: Divide the new first term \(\frac{15}{2}x\) by \(2x\) to get \(\frac{15}{4}\). Multiply the divisor \((2x - 3)\) by \(\frac{15}{4}\) to get \(\frac{15}{2}x - \frac{45}{4}\). Subtract this from \(\frac{15}{2}x + 2\):
So, \((2x^3 + 6x^2 - 6x + 2) \div (2x - 3) = x^2 + \frac{9}{2}x + \frac{15}{4} + \frac{53}{4(2x - 3)}\)
Problem 7: \((4x^3 - 6x^2 - 4x + 8) \div (2x - 1)\)
Step 1: Divide the first term of the dividend \(4x^3\) by the first term of the divisor \(2x\) to get \(2x^2\). Multiply the divisor \((2x - 1)\) by \(2x^2\) to get \(4x^3 - 2x^2\). Subtract this from the dividend:
Step 2: Divide the new first term \(-4x^2\) by \(2x\) to get \(-2x\). Multiply the divisor \((2x - 1)\) by \(-2x\) to get \(-4x^2 + 2x\). Subtract this from \(-4x^2 - 4x + 8\):
Step 3: Divide the new first term \(-6x\) by \(2x\) to get \(-3\). Multiply the divisor \((2x - 1)\) by \(-3\) to get \(-6x + 3\). Subtract this from \(-6x + 8\):
So, \((4x^3 - 6x^2 - 4x + 8) \div (2x - 1) = 2x^2 - 2x - 3 + \frac{5}{2x - 1}\)
Problem 8: \((2x^3 - 5x^2 + 3x - 6) \div (x + 4)\)
Step 1: Divide the first term of the dividend \(2x^3\) by the first term of the divisor \(x\) to get \(2x^2\). Multiply the divisor \((x + 4)\) by \(2x^2\) to get \(2x^3 + 8x^2\). Subtract this from the dividend:
Step 2: Divide the new first term \(-13x^2\) by \(x\) to get \(-13x\). Multiply the divisor \((x + 4)\) by \(-13x\) to get \(-13x^2 - 52x\). Subtract this from \(-13x^2 + 3x - 6\):
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Problem 5: \((3x^2 - 3x + 5) \div (x - 6)\)
Step 1: Use polynomial long division
Divide the first term of the dividend \(3x^2\) by the first term of the divisor \(x\) to get \(3x\). Multiply the divisor \((x - 6)\) by \(3x\) to get \(3x^2 - 18x\). Subtract this from the dividend:
Step 2: Divide the new first term \(15x\) by \(x\) to get \(15\). Multiply the divisor \((x - 6)\) by \(15\) to get \(15x - 90\). Subtract this from \(15x + 5\):
So, \((3x^2 - 3x + 5) \div (x - 6) = 3x + 15 + \frac{95}{x - 6}\)
Problem 6: \((2x^3 + 6x^2 - 6x + 2) \div (2x - 3)\)
Step 1: Divide the first term of the dividend \(2x^3\) by the first term of the divisor \(2x\) to get \(x^2\). Multiply the divisor \((2x - 3)\) by \(x^2\) to get \(2x^3 - 3x^2\). Subtract this from the dividend:
Step 2: Divide the new first term \(9x^2\) by \(2x\) to get \(\frac{9}{2}x\). Multiply the divisor \((2x - 3)\) by \(\frac{9}{2}x\) to get \(9x^2 - \frac{27}{2}x\). Subtract this from \(9x^2 - 6x + 2\):
Step 3: Divide the new first term \(\frac{15}{2}x\) by \(2x\) to get \(\frac{15}{4}\). Multiply the divisor \((2x - 3)\) by \(\frac{15}{4}\) to get \(\frac{15}{2}x - \frac{45}{4}\). Subtract this from \(\frac{15}{2}x + 2\):
So, \((2x^3 + 6x^2 - 6x + 2) \div (2x - 3) = x^2 + \frac{9}{2}x + \frac{15}{4} + \frac{53}{4(2x - 3)}\)
Problem 7: \((4x^3 - 6x^2 - 4x + 8) \div (2x - 1)\)
Step 1: Divide the first term of the dividend \(4x^3\) by the first term of the divisor \(2x\) to get \(2x^2\). Multiply the divisor \((2x - 1)\) by \(2x^2\) to get \(4x^3 - 2x^2\). Subtract this from the dividend:
Step 2: Divide the new first term \(-4x^2\) by \(2x\) to get \(-2x\). Multiply the divisor \((2x - 1)\) by \(-2x\) to get \(-4x^2 + 2x\). Subtract this from \(-4x^2 - 4x + 8\):
Step 3: Divide the new first term \(-6x\) by \(2x\) to get \(-3\). Multiply the divisor \((2x - 1)\) by \(-3\) to get \(-6x + 3\). Subtract this from \(-6x + 8\):
So, \((4x^3 - 6x^2 - 4x + 8) \div (2x - 1) = 2x^2 - 2x - 3 + \frac{5}{2x - 1}\)
Problem 8: \((2x^3 - 5x^2 + 3x - 6) \div (x + 4)\)
Step 1: Divide the first term of the dividend \(2x^3\) by the first term of the divisor \(x\) to get \(2x^2\). Multiply the divisor \((x + 4)\) by \(2x^2\) to get \(2x^3 + 8x^2\). Subtract this from the dividend:
Step 2: Divide the new first term \(-13x^2\) by \(x\) to get \(-13x\). Multiply the divisor \((x + 4)\) by \(-13x\) to get \(-13x^2 - 52x\). Subtract this from \(-13x^2 + 3x - 6\):
Step 3: Divide the new first term \(55x\) by \(x\) to get \(55\). Multiply the divisor \((x + 4)\) by \(55\) to get \(55x + 220\). Subtract this from \(55x - 6\):
So, \((2x^3 - 5x^2 + 3x - 6) \div (x + 4) = 2x^2 - 13x + 55 - \frac{226}{x + 4}\)
Final Answers:
- \(3x + 15 + \frac{95}{x - 6}\)
- \(x^2 + \frac{9}{2}x + \frac{15}{4} + \frac{53}{4(2x - 3)}\)
- \(2x^2 - 2x - 3 + \frac{5}{2x - 1}\)
- \(2x^2 - 13x + 55 - \frac{226}{x + 4}\)