To decompose the polynomial \( x^3 - 2x^2 - 5x + 6 \) into linear factors, given that \( x - 1 \) is a factor, we can use polynomial long division to divide the cubic polynomial by \( x - 1 \).
Step 1: Polynomial Long Division
We divide \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \).
- Divide the leading term: \( \frac{x^3}{x} = x^2 \).
- Multiply \( x^2 \) by \( x - 1 \): \[ x^2 \cdot (x - 1) = x^3 - x^2. \]
- Subtract: \[ \begin{align*} (x^3 - 2x^2) - (x^3 - x^2) &= -2x^2 + x^2 = -x^2. \end{align*} \]
- Bring down the next term: \[ -x^2 - 5x \Rightarrow -x^2 - 5x. \]
- Divide the leading term: \( \frac{-x^2}{x} = -x \).
- Multiply \( -x \) by \( x - 1 \): \[ -x \cdot (x - 1) = -x^2 + x. \]
- Subtract: \[ (-x^2 - 5x) - (-x^2 + x) = -5x - x = -6x. \]
- Bring down the next term: \[ -6x + 6. \]
- Divide the leading term: \( \frac{-6x}{x} = -6 \).
- Multiply \( -6 \) by \( x - 1 \): \[ -6 \cdot (x - 1) = -6x + 6. \]
- Subtract: \[ (-6x + 6) - (-6x + 6) = 0. \]
The division is complete, and we find that: \[ x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 - x - 6). \]
Step 2: Factor the Quadratic
Now we need to factor \( x^2 - x - 6 \).
To factor \( x^2 - x - 6 \), we can look for two numbers that multiply to \(-6\) (the constant term) and add to \(-1\) (the coefficient of \(x\)). These numbers are \(-3\) and \(2\).
So we can factor it as: \[ x^2 - x - 6 = (x - 3)(x + 2). \]
Step 3: Write the Fully Factored Form
Combining the factors, we find: \[ x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2). \]
Final Answer
The polynomial \( x^3 - 2x^2 - 5x + 6 \) is decomposed into linear factors as: \[ (x - 1)(x - 3)(x + 2). \]
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