1. **Problem:** Show that $(x-2)$ is a factor of $f(x) = x^3 - 7x + 6$ and factorise $f(x)$ completely. 2. **Step 1: Verify if $(x-2)$ is a factor using the Factor Theorem.** The Factor Theorem states that $(x - c)$ is a factor of $f(x)$ if and only if $f(c) = 0$. Calculate $f(2)$: $$f(2) = 2^3 - 7(2) + 6 = 8 - 14 + 6 = 0$$ Since $f(2) = 0$, $(x-2)$ is indeed a factor. 3. **Step 2: Perform polynomial division to factorise $f(x)$ by $(x-2)$.** Divide $f(x)$ by $(x-2)$: Using synthetic division: - Coefficients: 1 (for $x^3$), 0 (for $x^2$), -7 (for $x$), 6 (constant) - Bring down 1 - Multiply 1 by 2 = 2, add to 0 = 2 - Multiply 2 by 2 = 4, add to -7 = -3 - Multiply -3 by 2 = -6, add to 6 = 0 (remainder) So the quotient is $x^2 + 2x - 3$. 4. **Step 3: Factorise the quadratic $x^2 + 2x - 3$.** Find two numbers that multiply to $-3$ and add to $2$: these are $3$ and $-1$. So, $$x^2 + 2x - 3 = (x + 3)(x - 1)$$ 5. **Final factorisation:** $$f(x) = (x - 2)(x + 3)(x - 1)$$ This completes the factorisation of $f(x)$.