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what is the quotient of \\((x^3 - 3x^2 + 3x - 2) \\div (x^2 - x + 1)\\)…

Question

what is the quotient of \\((x^3 - 3x^2 + 3x - 2) \div (x^2 - x + 1)\\)?

\\(x - 2\\)
\\(x + 2\\)
\\(x - 4\\)
\\(x + 1\\)

Explanation:

Set up the polynomial long division

We need to divide the polynomial \(x^3 - 3x^2 + 3x - 2\) by \(x^2 - x + 1\).
The dividend is \(x^3 - 3x^2 + 3x - 2\) and the divisor is \(x^2 - x + 1\).

Divide the leading terms

Divide the leading term of the dividend, \(x^3\), by the leading term of the divisor, \(x^2\):

$$ \frac{x^3}{x^2} = x $$

This gives the first term of our quotient.

Multiply and subtract the first term

Multiply \(x\) by the divisor \(x^2 - x + 1\):

$$ x(x^2 - x + 1) = x^3 - x^2 + x $$

Subtract this result from the original dividend:

$$ (x^3 - 3x^2 + 3x - 2) - (x^3 - x^2 + x) = -2x^2 + 2x - 2 $$

Divide the new leading term

Divide the leading term of the remainder, \(-2x^2\), by the leading term of the divisor, \(x^2\):

$$ \frac{-2x^2}{x^2} = -2 $$

This gives the second term of our quotient.

Multiply and subtract the second term

Multiply \(-2\) by the divisor \(x^2 - x + 1\):

$$ -2(x^2 - x + 1) = -2x^2 + 2x - 2 $$

Subtract this result from the current remainder:

$$ (-2x^2 + 2x - 2) - (-2x^2 + 2x - 2) = 0 $$

Since the remainder is \(0\), the division is exact, and the quotient is \(x - 2\).

Answer:

  • (A) \(x - 2\) (Correct answer)
  • (B) \(x + 2\)
  • (C) \(x - 4\)
  • (D) \(x + 1\)