What are the additional linear factors of x^3-6x^2+11x-6 if x-3 is a factor?

a. (x+2)(x+1)
b. (x+2)(x-1)
c. (x-2)(x+1)
d. (x-2)(x-1)

1 answer

To find the additional linear factors of the polynomial \( x^3 - 6x^2 + 11x - 6 \) given that \( x - 3 \) is a factor, we can use polynomial long division to divide the cubic polynomial by \( x - 3 \).

Let's perform the polynomial long division of \( x^3 - 6x^2 + 11x - 6 \) by \( x - 3 \):

  1. Divide the leading term \( x^3 \) by \( x \) to get \( x^2 \).
  2. Multiply \( x^2 \) by \( x - 3 \) to get \( x^3 - 3x^2 \).
  3. Subtract \( (x^3 - 3x^2) \) from \( (x^3 - 6x^2) \) to get \( -3x^2 \).
  4. Bring down the next term to get \( -3x^2 + 11x \).
  5. Divide the leading term \( -3x^2 \) by \( x \) to get \( -3x \).
  6. Multiply \( -3x \) by \( x - 3 \) to get \( -3x^2 + 9x \).
  7. Subtract \( (-3x^2 + 9x) \) from \( (-3x^2 + 11x) \) to get \( 2x \).
  8. Bring down the next term to get \( 2x - 6 \).
  9. Divide the leading term \( 2x \) by \( x \) to get \( 2 \).
  10. Multiply \( 2 \) by \( x - 3 \) to get \( 2x - 6 \).
  11. Subtract \( (2x - 6) \) from \( (2x - 6) \) to get \( 0 \).

The result of the division is \( x^2 - 3x + 2 \). So we can express the polynomial as:

\[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x^2 - 3x + 2) \]

Next, we need to factor \( x^2 - 3x + 2 \):

\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

Thus, we have:

\[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x - 1)(x - 2) \]

The additional linear factors, therefore, are \( (x - 1) \) and \( (x - 2) \).

The correct answer choice for these factors is:

d. (x - 2)(x - 1)