To find the remaining linear factors of the polynomial \( x^4 - 13x^3 - x^2 + 13x \) given that \( (x + 1) \) and \( (x - 1) \) are factors, we can start by performing polynomial division.
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First, we can factor the polynomial: \[ x^4 - 13x^3 - x^2 + 13x = x(x^3 - 13x^2 - x + 13) \]
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Now, we need to perform synthetic division on \( x^3 - 13x^2 - x + 13 \) using the known roots \( x = -1 \) and \( x = 1 \).
Synthetic Division by \( x + 1 \):
Using \( x = -1 \): \[ \begin{array}{r|rrrr} -1 & 1 & -13 & -1 & 13 \ & & -1 & 14 & -13 \ \hline & 1 & -14 & 13 & 0 \ \end{array} \] This gives us a quotient of \( x^2 - 14x + 13 \).
Then we divide \( x^2 - 14x + 13 \) by \( x - 1 \):
Using \( x = 1 \): \[ \begin{array}{r|rr} 1 & 1 & -14 & 13 \ & & 1 & -13 \ \hline & 1 & -13 & 0 \ \end{array} \] This gives us a quotient of \( x - 13 \).
The result of the full factorization is: \[ x^4 - 13x^3 - x^2 + 13x = x(x + 1)(x - 1)(x - 13) \]
So, the remaining linear factor of the polynomial is \( (x - 13) \).
Thus, the remaining linear factors of the polynomial are \( (x + 1) \), \( (x - 1) \), and \( (x - 13) \).
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