1. **Problem (a):** Given that $f(x) = (x - a)^2 g(x)$, where $f(x)$ and $g(x)$ are polynomials, show that $(x - a)$ is a factor of $f'(x)$. 2. **Step 1:** Differentiate $f(x)$ using the product rule: $$f'(x) = \frac{d}{dx}[(x - a)^2 g(x)] = 2(x - a) g(x) + (x - a)^2 g'(x)$$ 3. **Step 2:** Factor out $(x - a)$ from $f'(x)$: $$f'(x) = (x - a) \left[ 2 g(x) + (x - a) g'(x) \right]$$ 4. Since $2 g(x) + (x - a) g'(x)$ is a polynomial, $(x - a)$ is a factor of $f'(x)$. --- 5. **Problem (b):** Given that $(x - 3)^2$ is a factor of $2x^3 - 4x^2 + p x + q$, find $p$ and $q$. 6. **Step 1:** Since $(x - 3)^2$ is a factor, the polynomial and its derivative must both be zero at $x=3$. 7. **Step 2:** Evaluate the polynomial at $x=3$: $$2(3)^3 - 4(3)^2 + p(3) + q = 0$$ $$2 \times 27 - 4 \times 9 + 3p + q = 0$$ $$54 - 36 + 3p + q = 0$$ $$18 + 3p + q = 0 \implies 3p + q = -18$$ 8. **Step 3:** Differentiate the polynomial: $$f'(x) = 6x^2 - 8x + p$$ 9. **Step 4:** Evaluate $f'(x)$ at $x=3$: $$6(3)^2 - 8(3) + p = 0$$ $$6 \times 9 - 24 + p = 0$$ $$54 - 24 + p = 0$$ $$30 + p = 0 \implies p = -30$$ 10. **Step 5:** Substitute $p = -30$ into $3p + q = -18$: $$3(-30) + q = -18$$ $$-90 + q = -18$$ $$q = 72$$ **Final answers:** $$p = -30, \quad q = 72$$