Use the Factor Theorem to determine whether \( x-2 \) is a factor of \( P(x)=x^{3}-3 x^{2}-4 x+ \) Specifically, evaluate \( P \) at the proper value, and then determine whether \( x-2 \) is a factor. \( \begin{array}{l}P(\square)=\square \\ x-2 \text { is a factor of } P(x) \\ x-2 \text { is not a factor of } P(x)\end{array} \)

Use the properties of logarithms to rewrite the expression. Simplify the res \( \log _{3} \frac{\sqrt{x} \sqrt[3]{y}}{w} \sqrt{z} \) Choose the correct answer. A. \( -\frac{1}{2} \log _{3} x-\frac{1}{3} \log _{3} y+5 \log _{3} w+\frac{1}{2} \log _{3} z \) B. \( \frac{1}{2} \log _{3} x+\frac{1}{3} \log _{3} y-5 \log _{3} w-\frac{1}{2} \log _{3} z \) C. \( { }^{2} \log _{3} x+3 \log _{3} y-5 \log _{3} w-2 \log _{3} z \) D. This cannot be simplified.

1 Graph \( y-2=\frac{2}{3}(x-3) \)

b. \( -0.5 p+\frac{1}{2} p-2.75+\frac{2}{3} \)

Write in exponential form. \[ \log _{6} 216=3 \] Write the equivalent exponential equation. \( \square \) (Type an equation. Type your answer using expc

Solve the equation by factoring. Select the correct solution. \( 6 x^{2}-5 x-4=0 \) \( \begin{array}{ll}\text { O } & \text { A) } x=-\frac{4}{3},-\frac{1}{2} \\ \text { O } & \text { B) } x=-\frac{1}{2}, \frac{4}{3} \\ \text { O } & \text { C) } x=-\frac{4}{3}, \frac{1}{2} \\ \text { - } & \text { D) } x=\frac{1}{2}, \frac{4}{3}\end{array} \)