Decompose the polynomial x^3−2x^2−5x+6 into its linear factors, given that x−1 is a factor. (1 point)

Answer 1

To decompose the polynomial \( x^3 - 2x^2 - 5x + 6 \) into its linear factors, given that \( x - 1 \) is a factor, we can perform polynomial long division or use synthetic division.

Let's start by using synthetic division. We will divide \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \):

1. Set up the synthetic division. The zero corresponding to \( x - 1 \) is \( 1 \).
2. Write the coefficients of the polynomial \( 1, -2, -5, 6 \).

The synthetic division process will look like this:

  1 |  1  -2  -5   6
    |      1  -1  -6
    -------------------
      1  -1  -6   0

Explanation of steps:

• Bring down the leading coefficient \( 1 \).
• Multiply \( 1 \) (the zero) by \( 1 \) (the number just brought down) and add it to \( -2 \) to get \( -1 \).
• Multiply \( 1 \) by \( -1 \) and add it to \( -5 \) to get \( -6 \).
• Multiply \( 1 \) by \( -6 \) and add it to \( 6 \) to get \( 0 \).

Since the remainder is \( 0 \), we have confirmed that \( x - 1 \) is indeed a factor. The result of the synthetic division gives us the quotient \( x^2 - x - 6 \).

Next, we need to factor \( x^2 - x - 6 \). To do this, we look for two numbers that multiply to \( -6 \) (the constant term) and add up to \( -1 \) (the coefficient of \( x \)). The numbers \( -3 \) and \( 2 \) satisfy this condition:

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

Now we can express the original polynomial in terms of its linear factors:

\[ x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2) \]

Thus, the complete factorization into linear factors is:

\[ \boxed{(x - 1)(x - 3)(x + 2)} \]