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 to divide the polynomial by \( x - 1 \).
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Perform the polynomial long division of \( x^3 - 2x^2 - 5x + 6 \) by \( x - 1 \).
\[ \begin{array}{r|rrr} & 1 & -1 & -5 & +6 \ x-1 & 1 & -2 & -5 & +6 \ & & 1 & -1 & -6 \ \hline & 1 & -1 & -6& 0 \ \end{array} \]
The result of this division is \( x^2 - x - 6 \), and the remainder is 0, confirming that \( x - 1 \) is indeed a factor.
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Next, we need to factor \( x^2 - x - 6 \). We can do this by finding 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.
Thus, we can factor \( x^2 - x - 6 \) as:
\[ (x - 3)(x + 2) \]
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Now we can express the original polynomial \( x^3 - 2x^2 - 5x + 6 \) in factored form:
\[ x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2) \]
Therefore, the complete factorization of the polynomial \( x^3 - 2x^2 - 5x + 6 \) into its linear factors is:
\[ (x - 1)(x - 3)(x + 2) \]
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