Divide x^3 - 6x^2 + 11x - 6 by (x - 1).

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Multiple Choice

Divide x^3 - 6x^2 + 11x - 6 by (x - 1).

Explanation:
When dividing by a binomial of the form (x − a), you can use the Remainder Theorem: the remainder is P(a). If we plug in a = 1 into the polynomial P(x) = x^3 − 6x^2 + 11x − 6, we get P(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor and the division has no remainder. Using synthetic division with 1 as the root, the coefficients 1, −6, 11, −6 drop down to give the quotient coefficients: 1, −5, 6. Therefore the quotient is x^2 − 5x + 6. Checking by multiplying back, (x − 1)(x^2 − 5x + 6) expands to x^3 − 6x^2 + 11x − 6, which matches the original polynomial.

When dividing by a binomial of the form (x − a), you can use the Remainder Theorem: the remainder is P(a). If we plug in a = 1 into the polynomial P(x) = x^3 − 6x^2 + 11x − 6, we get P(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor and the division has no remainder.

Using synthetic division with 1 as the root, the coefficients 1, −6, 11, −6 drop down to give the quotient coefficients: 1, −5, 6. Therefore the quotient is x^2 − 5x + 6. Checking by multiplying back, (x − 1)(x^2 − 5x + 6) expands to x^3 − 6x^2 + 11x − 6, which matches the original polynomial.

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