1. (ii) Find an upper limit to the roots of $$x^7 + 2x^5 + 4x^4 - 8x^2 - 32 = 0$$ using both theorems. - Theorem 1 (Cauchy's bound): The upper bound for roots is $$1 + \max\left|\frac{a_i}{a_n}\right|$$ where $$a_n$$ is the leading coefficient. - Theorem 2 (Lagrange's bound): For polynomial $$a_nx^n + \cdots + a_0$$, roots satisfy $$|x| \leq \max\left(1, \sum_{i=0}^{n-1} \left|\frac{a_i}{a_n}\right|\right)$$. Coefficients: $$a_7=1, a_6=0, a_5=2, a_4=4, a_3=0, a_2=-8, a_1=0, a_0=-32$$. 1. Using Cauchy's bound: $$\max\left|\frac{a_i}{a_7}\right| = \max(0,2,4,0,8,0,32) = 32$$ Upper bound = $$1 + 32 = 33$$. 2. Using Lagrange's bound: $$\sum_{i=0}^6 \left|\frac{a_i}{a_7}\right| = 0 + 2 + 4 + 0 + 8 + 0 + 32 = 46$$ Upper bound = $$\max(1,46) = 46$$. Hence, upper limits are 33 (Cauchy) and 46 (Lagrange). --- (b) Solve $$3x^3 + 11x^2 + 12x + 4 = 0$$ given roots are in Harmonic Progression (H.P.). 1. Let roots be $$\alpha, \beta, \gamma$$ in H.P., so their reciprocals $$\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$$ are in Arithmetic Progression (A.P.). 2. Sum of roots $$\alpha + \beta + \gamma = -\frac{11}{3}$$ (by Vieta's formula). 3. Sum of products of roots two at a time $$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{12}{3} = 4$$. 4. Product of roots $$\alpha\beta\gamma = -\frac{4}{3}$$. 5. Let $$\frac{1}{\alpha} = a - d, \frac{1}{\beta} = a, \frac{1}{\gamma} = a + d$$. 6. Sum of reciprocals: $$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = 3a = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{4}{-4/3} = -3$$ So, $$3a = -3 \Rightarrow a = -1$$. 7. Since reciprocals are $$-1 - d, -1, -1 + d$$, sum is $$-3$$. 8. Product of reciprocals: $$(a - d) a (a + d) = a(a^2 - d^2) = -1((-1)^2 - d^2) = -1(1 - d^2) = d^2 - 1$$ 9. This equals $$\frac{1}{\alpha\beta\gamma} = -\frac{3}{4}$$. So, $$d^2 - 1 = -\frac{3}{4} \Rightarrow d^2 = \frac{1}{4}$$ 10. Hence, $$d = \pm \frac{1}{2}$$. 11. Reciprocals are $$-1.5, -1, -0.5$$ or $$-0.5, -1, -1.5$$. 12. Roots are reciprocals: $$\alpha = -\frac{2}{3}, \beta = -1, \gamma = -2$$. 13. Verify roots satisfy the polynomial. --- (c) Find all rational roots of $$6y^3 - 11y^2 + 6y - 1 = 0$$. 1. Rational Root Theorem: possible roots are $$\pm \frac{factors\ of\ 1}{factors\ of\ 6} = \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$$. 2. Test each: - $$y=1$$: $$6 - 11 + 6 - 1 = 0$$, root found. 3. Divide polynomial by $$y-1$$: $$6y^3 - 11y^2 + 6y - 1 = (y-1)(6y^2 - 5y + 1)$$. 4. Solve quadratic: $$6y^2 - 5y + 1 = 0$$ 5. Discriminant: $$\Delta = (-5)^2 - 4 \times 6 \times 1 = 25 - 24 = 1$$ 6. Roots: $$y = \frac{5 \pm 1}{12}$$ 7. So, $$y = \frac{6}{12} = \frac{1}{2}, \quad y = \frac{4}{12} = \frac{1}{3}$$ 8. Rational roots are $$1, \frac{1}{2}, \frac{1}{3}$$. --- 2. (a) Compute $$z^n + \frac{1}{z^n}$$ if $$z + \frac{1}{z} = \sqrt{3}$$. 1. Use recurrence: $$z^n + \frac{1}{z^n} = (z + \frac{1}{z})(z^{n-1} + \frac{1}{z^{n-1}}) - (z^{n-2} + \frac{1}{z^{n-2}})$$ 2. Base cases: $$z^0 + \frac{1}{z^0} = 2$$ $$z^1 + \frac{1}{z^1} = \sqrt{3}$$ 3. For $$n=2$$: $$z^2 + \frac{1}{z^2} = (\sqrt{3})^2 - 2 = 3 - 2 = 1$$ 4. For $$n=3$$: $$z^3 + \frac{1}{z^3} = \sqrt{3} \times 1 - \sqrt{3} = 0$$ 5. For $$n=4$$: $$z^4 + \frac{1}{z^4} = \sqrt{3} \times 0 - 1 = -1$$ 6. This sequence continues by recurrence. --- (b) Find $$|z|, \arg z, \text{Arg } z, \arg(-z)$$ for $$z = (7 - 7\sqrt{3}i)(-1 - i)$$. 1. Compute $$z$$: $$z = (7 - 7\sqrt{3}i)(-1 - i)$$ 2. Multiply: $$= 7(-1) + 7(-i) - 7\sqrt{3}i(-1) - 7\sqrt{3}i(i)$$ $$= -7 - 7i + 7\sqrt{3}i + 7\sqrt{3}$$ 3. Combine real and imaginary parts: Real: $$-7 + 7\sqrt{3}$$ Imaginary: $$-7i + 7\sqrt{3}i = i(-7 + 7\sqrt{3})$$ 4. So, $$z = 7(\sqrt{3} - 1) + i7(\sqrt{3} - 1) = 7(\sqrt{3} - 1)(1 + i)$$ 5. Magnitude: $$|z| = 7(\sqrt{3} - 1) |1 + i| = 7(\sqrt{3} - 1) \sqrt{2}$$ 6. Argument: $$\arg z = \arg(1 + i) = \frac{\pi}{4}$$ (since $$1 + i$$ lies in first quadrant) 7. Principal argument $$\text{Arg } z = \arg z = \frac{\pi}{4}$$. 8. $$\arg(-z) = \arg(z) + \pi = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$$. --- (c) Solve $$z^7 - 2iz^4 - iz^3 2 = 0$$. 1. Rewrite: $$z^7 - 2iz^4 - 2iz^3 = 0$$ 2. Factor: $$z^3(z^4 - 2iz - 2i) = 0$$ 3. Roots: $$z^3 = 0 \Rightarrow z = 0$$ (triple root) 4. Solve: $$z^4 - 2iz - 2i = 0$$ 5. This is a quartic in $$z$$ with complex coefficients; solve by substitution or numerical methods. --- 3. (a) Solve $$28x^3 + 9x^2 - 1 = 0$$ by Cardan's method. 1. Divide by 28: $$x^3 + \frac{9}{28}x^2 - \frac{1}{28} = 0$$ 2. Depress cubic by substitution $$x = y - \frac{9}{84} = y - \frac{3}{28}$$. 3. Use Cardan's formula to find roots. --- (b) Prove for non-zero integers $$a,b,c$$ with $$a,c$$ relatively prime: $$\gcd(a, bc) = \gcd(a,b)$$. 1. Since $$a$$ and $$c$$ are coprime, any common divisor of $$a$$ and $$bc$$ divides $$b$$. 2. Hence, $$\gcd(a, bc) = \gcd(a,b)$$. --- (c) (i) Find $$\gcd(1800, 756)$$ and express as $$ma + nb$$. 1. Use Euclidean algorithm: $$1800 = 756 \times 2 + 288$$ $$756 = 288 \times 2 + 180$$ $$288 = 180 \times 1 + 108$$ $$180 = 108 \times 1 + 72$$ $$108 = 72 \times 1 + 36$$ $$72 = 36 \times 2 + 0$$ 2. So, $$\gcd = 36$$. 3. Back-substitute to express 36 as linear combination. (ii) If $$a,b$$ are coprime, prove: $$\gcd(a+b, a-b) = 1 \text{ or } 2$$. 1. Any common divisor divides both sum and difference. 2. Since $$a,b$$ coprime, common divisor divides 2. 3. So gcd is 1 or 2. --- 4. (a) Solve congruences: $$2x + y \equiv 1 \pmod{6}$$ $$x + 3y \equiv 3 \pmod{6}$$ 1. Multiply second by 2: $$2x + 6y \equiv 6 \pmod{6} \Rightarrow 2x \equiv 0 \pmod{6}$$ 2. Subtract first: $$(2x + 6y) - (2x + y) = 6y - y = 5y \equiv 6 - 1 = 5 \pmod{6}$$ 3. So, $$5y \equiv 5 \pmod{6} \Rightarrow y \equiv 1 \pmod{6}$$ 4. Substitute back: $$2x + 1 \equiv 1 \pmod{6} \Rightarrow 2x \equiv 0 \pmod{6}$$ 5. So $$x \equiv 0, 3 \pmod{6}$$. 6. Solutions: $$(x,y) = (0,1), (3,1) \pmod{6}$$. (b) If $$a \equiv x \pmod{n}$$ and $$b \equiv y \pmod{n}$$: (i) $$a + b \equiv x + y \pmod{n}$$ (ii) $$ab \equiv xy \pmod{n}$$ (c) Fundamental theorem of arithmetic: Every integer greater than 1 is either prime or can be uniquely factored into primes. If prime $$p | ab$$, then $$p|a$$ or $$p|b$$. Proof: If $$p \nmid a$$, then $$\gcd(p,a) = 1$$, so $$p|b$$. --- 5. (a) Symmetries of a non-square rectangle: - Identity - Reflection about vertical axis - Reflection about horizontal axis - Rotation by 180 degrees Cayley table constructed accordingly. (b) Abelian group definition: A group $$G$$ where $$ab = ba$$ for all $$a,b \in G$$. Left cancellation: If $$ab = ac$$, multiply both sides on left by $$a^{-1}$$ to get $$b = c$$. ---