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Divide. If there is a remainder, include it as a simplified fraction.\newline(2c2+5c7)÷(c1)(2c^2 + 5c - 7) \div (c - 1)

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Q. Divide. If there is a remainder, include it as a simplified fraction.\newline(2c2+5c7)÷(c1)(2c^2 + 5c - 7) \div (c - 1)
  1. Divide by First Term: We will use polynomial long division to divide (2c2+5c7)(2c^2 + 5c - 7) by (c1)(c - 1). First, we divide the first term of the dividend, 2c22c^2, by the first term of the divisor, cc, to get the first term of the quotient. 2c2÷c=2c2c^2 \div c = 2c
  2. Multiply and Subtract: Now, we multiply the divisor c1c - 1 by the first term of the quotient 2c2c and subtract the result from the dividend.(2c)(c1)=2c22c(2c)(c - 1) = 2c^2 - 2cSubtract this from the dividend:(2c2+5c7)(2c22c)=5c2c7=3c7(2c^2 + 5c - 7) - (2c^2 - 2c) = 5c - 2c - 7 = 3c - 7
  3. Divide New Term: Next, we divide the new first term of the remaining polynomial, 3c3c, by the first term of the divisor, cc. \newline3c÷c=33c \div c = 3
  4. Multiply and Subtract: We multiply the divisor (c1)(c - 1) by the new term of the quotient (3)(3) and subtract the result from the remaining polynomial.\newline(3)(c1)=3c3(3)(c - 1) = 3c - 3\newlineSubtract this from the remaining polynomial:\newline(3c7)(3c3)=7+3=4(3c - 7) - (3c - 3) = -7 + 3 = -4
  5. Final Result: Since 4-4 is a constant and the divisor is a linear term (c1)(c - 1), we cannot divide further. Therefore, 4-4 is the remainder.\newlineThe quotient is 2c+32c + 3 with a remainder of 4-4.

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