1. **State the problem:** We have a function $f(x) = (x - 5)(x + a) - bx$ where $a > 0$ and constants $a, b$. When $f(x)$ is divided by $x - 5$, the remainder is 80. (a) Find $b$. (b) Given $f(x - 1) = (x - 2)(sx^2 + tx + u)$ for some constants $s, t, u$. (i) Find $a$. (ii) Determine if all roots of $f(x) = 0$ are rational and explain. 2. **Use the Remainder Theorem:** When a polynomial $f(x)$ is divided by $x - c$, the remainder is $f(c)$. 3. **Find $b$ using the remainder condition:** Calculate $f(5)$: $$f(5) = (5 - 5)(5 + a) - b \times 5 = 0 - 5b = -5b$$ Given remainder is 80, so: $$-5b = 80 \implies b = -\frac{80}{5} = -16$$ 4. **Rewrite $f(x)$ with $b = -16$:** $$f(x) = (x - 5)(x + a) - (-16)x = (x - 5)(x + a) + 16x$$ Expand: $$= x^2 + ax - 5x - 5a + 16x = x^2 + (a + 11)x - 5a$$ 5. **Use the given factorization for $f(x-1)$:** $$f(x - 1) = (x - 2)(sx^2 + tx + u)$$ Substitute $x - 1$ into $f(x)$: $$f(x - 1) = (x - 1)^2 + (a + 11)(x - 1) - 5a$$ Expand: $$(x - 1)^2 = x^2 - 2x + 1$$ So: $$f(x - 1) = x^2 - 2x + 1 + (a + 11)x - (a + 11) - 5a = x^2 - 2x + 1 + (a + 11)x - a - 11 - 5a$$ Simplify: $$= x^2 + (a + 9)x + (1 - 6a - 11) = x^2 + (a + 9)x - 6a - 10$$ 6. **Express $f(x - 1)$ as $(x - 2)(sx^2 + tx + u)$:** Multiply out: $$(x - 2)(sx^2 + tx + u) = sx^3 + tx^2 + ux - 2sx^2 - 2tx - 2u = sx^3 + (t - 2s)x^2 + (u - 2t)x - 2u$$ 7. **Match degrees:** $f(x - 1)$ is quadratic, so $s = 0$ to eliminate $x^3$ term. Then: $$(x - 2)(tx + u) = t x^2 + u x - 2 t x - 2 u = t x^2 + (u - 2 t) x - 2 u$$ 8. **Match coefficients with $f(x - 1) = x^2 + (a + 9)x - 6a - 10$:** Equate coefficients: - Coefficient of $x^2$: $t = 1$ - Coefficient of $x$: $u - 2 t = a + 9$ - Constant term: $-2 u = -6 a - 10$ 9. **Solve for $u$ and $a$:** From constant term: $$-2 u = -6 a - 10 \implies u = 3 a + 5$$ From $x$ coefficient: $$u - 2 = a + 9$$ Substitute $u$: $$3 a + 5 - 2 = a + 9 \implies 3 a + 3 = a + 9 \implies 2 a = 6 \implies a = 3$$ 10. **Find $u$:** $$u = 3(3) + 5 = 9 + 5 = 14$$ 11. **Check $b$ and $a$ values:** We have $a = 3$, $b = -16$. 12. **Check if roots of $f(x) = 0$ are rational:** Recall: $$f(x) = x^2 + (a + 11) x - 5 a = x^2 + 14 x - 15$$ Calculate discriminant: $$\Delta = 14^2 - 4 \times 1 \times (-15) = 196 + 60 = 256$$ Since $\sqrt{256} = 16$ is an integer, roots are: $$x = \frac{-14 \pm 16}{2}$$ Roots: $$x_1 = \frac{-14 + 16}{2} = 1$$ $$x_2 = \frac{-14 - 16}{2} = -15$$ Both roots are rational. **Final answers:** (a) $b = -16$ (b)(i) $a = 3$ (b)(ii) Yes, all roots of $f(x) = 0$ are rational because the discriminant is a perfect square and roots are integers.