6. Determine whether $x-1$ is a factor of the polynomial $P(x) = x^3 - 4x^2 + x + 6$. Step 1: Apply the Factor Theorem. According to the Factor Theorem, $(x-c)$ is a factor of a polynomial $P(x)$ if and only if $P(c) = 0$. In this case, $c=1$. Step 2: Evaluate $P(1)$. $$ P(1) = (1)^3 - 4(1)^2 + (1) + 6 $$ $$ P(1) = 1 - 4(1) + 1 + 6 $$ $$ P(1) = 1 - 4 + 1 + 6 $$ $$ P(1) = -3 + 1 + 6 $$ $$ P(1) = -2 + 6 $$ $$ P(1) = 4 $$ Step 3: Conclude based on the result. Since $P(1) = 4 \neq 0$, $x-1$ is not a factor of $P(x)$. 7. Given that $x+2$ is a factor of the polynomial $P(x) = x^3 + x^2 - 4x - 4$, find the complete factorization of $P(x)$. Step 1: Use synthetic division to divide $P(x)$ by $(x+2)$. Since $(x+2)$ is a factor, we use the root $x=-2$ for synthetic division. $$ \begin{array}{c|cccc} -2 & 1 & 1 & -4 & -4 \\ & & -2 & 2 & 4 \\ \hline & 1 & -1 & -2 & 0 \\ \end{array} $$ The remainder is $0$, confirming that $(x+2)$ is a factor. The quotient is $x^2 - x - 2$. Step 2: Factor the quadratic quotient. We need to factor $x^2 - x - 2$. We look for two numbers that multiply to $-2$ and add to $-1$. These numbers are $-2$ and $1$. $$ x^2 - x - 2 = (x-2)(x+1) $$ Step 3: Write the complete factorization of $P(x)$. $$ P(x) = (x+2)(x^2 - x - 2) $$ $$ P(x) = \boxed{(x+2)(x-2)(x+1)} $$ 8. Explain why the following expression is not a polynomial: $f(x) = \frac{3}{x^2} + x$. A polynomial is an expression where all variables have non-negative integer exponents. The given expression can be rewritten as $f(x) = 3x^{-2} + x^1$. The term $3x^{-2}$ has an exponent of $-2$ for the variable $x$. Since $-2$ is a negative integer, it violates the definition of a polynomial.