1. **State the problem:** We are given roots $\alpha$ and $\beta$ of the quadratic equation $x^2 + 3x - 5 = 0$ with conditions $\alpha^2 + \beta^2 = 19$, $\alpha - \beta = \sqrt{29}$, and $\alpha > \beta$. We need to factorise $\alpha^4 - \beta^4$ completely. 2. **Recall formulas and rules:** - Sum and product of roots for $x^2 + 3x - 5 = 0$ are $\alpha + \beta = -3$ and $\alpha \beta = -5$. - Difference of squares: $a^2 - b^2 = (a-b)(a+b)$. - Difference of fourth powers: $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$. 3. **Express $\alpha^4 - \beta^4$ using difference of squares:** $$\alpha^4 - \beta^4 = (\alpha^2 - \beta^2)(\alpha^2 + \beta^2)$$ 4. **Calculate $\alpha^2 - \beta^2$:** $$\alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta)$$ Given $\alpha - \beta = \sqrt{29}$ and $\alpha + \beta = -3$, so $$\alpha^2 - \beta^2 = \sqrt{29} \times (-3) = -3\sqrt{29}$$ 5. **Use given $\alpha^2 + \beta^2 = 19$** 6. **Substitute values back:** $$\alpha^4 - \beta^4 = (-3\sqrt{29})(19) = -57\sqrt{29}$$ 7. **Factorise completely:** Recall from step 3: $$\alpha^4 - \beta^4 = (\alpha^2 - \beta^2)(\alpha^2 + \beta^2)$$ We can factor $\alpha^2 - \beta^2$ further: $$\alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta)$$ So the complete factorisation is: $$\alpha^4 - \beta^4 = (\alpha - \beta)(\alpha + \beta)(\alpha^2 + \beta^2)$$ **Final answer:** $$\boxed{(\alpha - \beta)(\alpha + \beta)(\alpha^2 + \beta^2)}$$ Substituting the known values: $$= (\sqrt{29})(-3)(19) = -57\sqrt{29}$$