1. **Problem statement:** Find the remainder when $f(x) = 3x^3 - 14x^2 - 47x - 14$ is divided by $(x - 3)$. 2. **Formula used:** The Remainder Theorem states that the remainder when a polynomial $f(x)$ is divided by $(x - a)$ is $f(a)$. 3. **Apply the theorem:** Calculate $f(3)$: $$f(3) = 3(3)^3 - 14(3)^2 - 47(3) - 14$$ $$= 3 \times 27 - 14 \times 9 - 141 - 14$$ $$= 81 - 126 - 141 - 14$$ 4. **Simplify step-by-step:** $$81 - 126 = -45$$ $$-45 - 141 = -186$$ $$-186 - 14 = -200$$ 5. **Final answer:** The remainder when $f(x)$ is divided by $(x - 3)$ is $\boxed{-200}$. 1. **Problem statement:** Given that $(x + 2)$ is a factor of $f(x)$, factorise $f(x)$ completely. 2. **Step 1: Write $f(x)$ as $(x + 2)(ax^2 + bx + c)$:** We want to find $a$, $b$, and $c$ such that: $$f(x) = (x + 2)(ax^2 + bx + c)$$ 3. **Expand the right side:** $$ (x + 2)(ax^2 + bx + c) = ax^3 + bx^2 + cx + 2ax^2 + 2bx + 2c $$ $$= ax^3 + (b + 2a)x^2 + (c + 2b)x + 2c$$ 4. **Match coefficients with $f(x) = 3x^3 - 14x^2 - 47x - 14$:** - Coefficient of $x^3$: $a = 3$ - Coefficient of $x^2$: $b + 2a = -14$ - Coefficient of $x$: $c + 2b = -47$ - Constant term: $2c = -14$ 5. **Solve for $c$:** $$2c = -14 \implies c = -7$$ 6. **Solve for $b$:** $$b + 2(3) = -14 \implies b + 6 = -14 \implies b = -20$$ 7. **Check $c + 2b = -47$:** $$-7 + 2(-20) = -7 - 40 = -47$$ Correct. 8. **Write factorisation:** $$f(x) = (x + 2)(3x^2 - 20x - 7)$$ 9. **Factorise quadratic $3x^2 - 20x - 7$:** Find two numbers that multiply to $3 \times (-7) = -21$ and add to $-20$. These are $-21$ and $1$. 10. **Split middle term:** $$3x^2 - 21x + x - 7$$ 11. **Factor by grouping:** $$3x(x - 7) + 1(x - 7) = (3x + 1)(x - 7)$$ 12. **Complete factorisation:** $$f(x) = (x + 2)(3x + 1)(x - 7)$$ **Final answers:** - Remainder when divided by $(x - 3)$ is $-200$. - Complete factorisation is $f(x) = (x + 2)(3x + 1)(x - 7)$.