Here are the solutions to the questions: 2.1 Prove that $(2x-1)$ is a factor of $f(x) = 12x^3 - 28x^2 - 3x + 7$. Step 1: Find the root of the factor $(2x-1)$. Set $2x-1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$. Step 2: Substitute $x = \frac{1}{2}$ into the polynomial $f(x)$. $$f\left(\frac{1}{2}\right) = 12\left(\frac{1}{2}\right)^3 - 28\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 7$$ Step 3: Calculate the value of $f\left(\frac{1}{2}\right)$. $$f\left(\frac{1}{2}\right) = 12\left(\frac{1}{8}\right) - 28\left(\frac{1}{4}\right) - \frac{3}{2} + 7$$ $$f\left(\frac{1}{2}\right) = \frac{12}{8} - \frac{28}{4} - \frac{3}{2} + 7$$ $$f\left(\frac{1}{2}\right) = \frac{3}{2} - 7 - \frac{3}{2} + 7$$ $$f\left(\frac{1}{2}\right) = 0$$ Step 4: Conclude using the Factor Theorem. Since $f\left(\frac{1}{2}\right) = 0$, by the Factor Theorem, $(2x-1)$ is a factor of $f(x)$. 2.2 The expression $x^3 - ax^2 + ax + b$ is divisible by $(x+1)$, $(x-3)$ and $(x+m)$. Let $P(x) = x^3 - ax^2 + ax + b$. 2.2.1 Determine the values of $a$ and $b$. Step 1: Use the Factor Theorem for $(x+1)$. Since $P(x)$ is divisible by $(x+1)$, $P(-1) = 0$. $$P(-1) = (-1)^3 - a(-1)^2 + a(-1) + b = 0$$ $$-1 - a - a + b = 0$$ $$-1 - 2a + b = 0 \implies b = 2a + 1 \quad \text{(Equation 1)}$$ Step 2: Use the Factor Theorem for $(x-3)$. Since $P(x)$ is divisible by $(