The polynomial p(x)=x^(3)-7x-6 has a known factor of (x+1). Rewrite p(x) as a product of linear factors. p(x)=

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The polynomial  p(x)=x^(3)-7x-6 has a known factor of  (x+1). Rewrite  p(x) as a product of linear factors.  p(x)=

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Factor Finding Method: Since we know that (x + 1) is a factor of p(x), we can perform polynomial division or use synthetic division to divide p(x) by (x + 1) to find the other factors.Synthetic Division Setup: Let's use synthetic division to divide p(x) by (x + 1). We set up the synthetic division with -1 (the zero of the factor x + 1) and the coefficients of p(x): 1 (for x^3), 0 (for x^2, since there is no x^2 term), -7 (for x), and -6 (constant term).Performing Synthetic Division: Performing synthetic division, we bring down the leading 1, multiply it by -1 to get -1, add this to the next coefficient (0) to get -1, multiply -1 by -1 to get 1, add this to -7 to get -6, multiply -6 by -1 to get 6, and add this to -6 to get 0. The result of the synthetic division is the coefficients of the quotient polynomial: 1 (for x^2), -1 (for x), and -6 (for the constant term).Quotient Polynomial: The quotient polynomial is x^2 - x - 6. We can factor this quadratic polynomial to find the other linear factors of p(x).Factoring Quadratic Polynomial: Factoring x^2 - x - 6, we look for two numbers that multiply to -6 and add to -1. These numbers are -3 and 2. Therefore, x^2 - x - 6 factors into (x - 3)(x + 2).Final Solution: Now we can write p(x) as a product of its linear factors: p(x) = (x + 1)(x - 3)(x + 2).

The polynomial $p(x)=x^{3}-7 x-6$ has a known factor of $(x+1)$ . $\newline$ Rewrite $p(x)$ as a product of linear factors. $\newline$ $p(x)=$

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The polynomial p(x)=x3−7x−6 p(x)=x^{3}-7 x-6 p(x)=x3−7x−6 has a known factor of (x+1) (x+1) (x+1).\newlineRewrite p(x) p(x) p(x) as a product of linear factors.\newlinep(x)= p(x)= p(x)=

The polynomial $p(x)=x^{3}-7 x-6$ has a known factor of $(x+1)$ . $\newline$ Rewrite $p(x)$ as a product of linear factors. $\newline$ $p(x)=$