The polynomial $$ x^{4}-6 x^{2}+x+a $$ is denoted by $$ p(x) $$. It is given that $$ x+1 $$ is a factor of $$ p(x) $$. Find the value of $$ a $$. When $$ a $$ has this value, verify that $$ x-2 $$ is also a factor of $$ p(x) $$, and hence factorise $$ p(x) $$ completely.

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The value of $$ a $$ is 6. The complete factorization of $$ p(x) $$ is $$ p(x) = (x + 1)(x - 2)\left(x - \frac{-1 + \sqrt{13}}{2}\right)\left(x - \frac{-1 - \sqrt{13}}{2}\right) $$.

The polynomial \( x^{4}-6 x^{2}+x+a \) is denoted by \( p(x) \). It is given that \( x+1 \) is a factor of \( p(x) \). Find the value of \( a \). When \( a \) has this value, verify that \( x-2 \) is also a factor of \( p(x) \), and hence factorise \( p(x) \) completely.

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The polynomial \( x^{4}-6 x^{2}+x+a \) is denoted by \( p(x) \). It is given that \( x+1 \) is a factor of \( p(x) \). Find the value of \( a \). When \( a \) has this value, verify that \( x-2 \) is also a factor of \( p(x) \), and hence factorise \( p(x) \) completely.

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