1. **Problem statement:** (a) Factorise completely the expression $x^3 - 4x$. (b) Sketch the curve with equation $y = x^3 - 4x$, showing the coordinates of the points where the curve crosses the x-axis. (c) Sketch the curve with equation $y = (x-1)^3 - 4(x-1)$, showing the coordinates of the points where the curve crosses the x-axis. 2. **Formula and rules:** To factorise expressions like $x^3 - 4x$, first look for common factors and then apply factorisation formulas such as difference of squares or sum/difference of cubes if applicable. 3. **Step (a) Factorisation:** - Start with $x^3 - 4x$. - Factor out the common factor $x$: $$x^3 - 4x = x(x^2 - 4)$$ - Recognize $x^2 - 4$ as a difference of squares: $$x^2 - 4 = (x - 2)(x + 2)$$ - So the complete factorisation is: $$x^3 - 4x = x(x - 2)(x + 2)$$ 4. **Step (b) Sketch $y = x^3 - 4x$ and find x-intercepts:** - The x-intercepts occur where $y=0$: $$x(x - 2)(x + 2) = 0$$ - So $x=0$, $x=2$, or $x=-2$. - Coordinates of x-intercepts are $(0,0)$, $(2,0)$, and $(-2,0)$. 5. **Step (c) Sketch $y = (x-1)^3 - 4(x-1)$ and find x-intercepts:** - Let $u = x - 1$, then: $$y = u^3 - 4u$$ - Factorise as before: $$y = u(u^2 - 4) = u(u - 2)(u + 2)$$ - Set $y=0$ to find x-intercepts: $$u=0, u=2, u=-2$$ - Substitute back $u = x - 1$: $$x - 1 = 0 ightarrow x=1$$ $$x - 1 = 2 ightarrow x=3$$ $$x - 1 = -2 ightarrow x=-1$$ - Coordinates of x-intercepts are $(1,0)$, $(3,0)$, and $(-1,0)$. **Final answers:** (a) $x^3 - 4x = x(x - 2)(x + 2)$ (b) x-intercepts at $(0,0)$, $(2,0)$, $(-2,0)$ (c) x-intercepts at $(1,0)$, $(3,0)$, $(-1,0)$