1. **Problem 6:** The polynomial is $3x^3 + 8x^2 - 15x + k$ and $(x - 1)$ is a factor. Find $k$ and factorise the polynomial completely. Step 1: Since $(x - 1)$ is a factor, substitute $x=1$ into the polynomial and set equal to zero: $$3(1)^3 + 8(1)^2 - 15(1) + k = 0$$ $$3 + 8 - 15 + k = 0$$ $$-4 + k = 0$$ Step 2: Solve for $k$: $$k = 4$$ Step 3: The polynomial becomes: $$3x^3 + 8x^2 - 15x + 4$$ Step 4: Use polynomial division or factor by grouping to factorise: Divide by $(x-1)$: Using synthetic division: Coefficients: 3, 8, -15, 4 Bring down 3. Multiply 3 by 1 = 3; add to 8 = 11. Multiply 11 by 1 = 11; add to -15 = -4. Multiply -4 by 1 = -4; add to 4 = 0. Quotient polynomial: $3x^2 + 11x - 4$ Step 5: Factor $3x^2 + 11x - 4$: Find two numbers that multiply to $3 imes (-4) = -12$ and add to 11: 12 and -1. Rewrite: $$3x^2 + 12x - x - 4$$ Group: $$(3x^2 + 12x) - (x + 4)$$ Factor: $$3x(x + 4) - 1(x + 4) = (3x - 1)(x + 4)$$ Step 6: Final factorisation: $$(x - 1)(3x - 1)(x + 4)$$ --- 2. **Problem 7:** Given $x, y, z$ are in continued proportion, prove: $$\frac{x}{y^2 z^2} + \frac{y}{z^2 x^2} + \frac{z}{x^2 y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$$ Step 1: Since $x, y, z$ are in continued proportion, there exists $r$ such that: $$y = xr, \quad z = xr^2$$ Step 2: Substitute into left side: $$\frac{x}{(xr)^2 (xr^2)^2} + \frac{xr}{(xr^2)^2 (x)^2} + \frac{xr^2}{(x)^2 (xr)^2}$$ Calculate denominators: $$(xr)^2 = x^2 r^2$$ $$(xr^2)^2 = x^2 r^4$$ So first term denominator: $$x^2 r^2 \times x^2 r^4 = x^4 r^{6}$$ First term: $$\frac{x}{x^4 r^6} = \frac{1}{x^3 r^6}$$ Second term denominator: $$(xr^2)^2 (x)^2 = x^2 r^4 \times x^2 = x^4 r^4$$ Second term: $$\frac{xr}{x^4 r^4} = \frac{r}{x^3 r^4} = \frac{1}{x^3 r^3}$$ Third term denominator: $$(x)^2 (xr)^2 = x^2 \times x^2 r^2 = x^4 r^2$$ Third term: $$\frac{xr^2}{x^4 r^2} = \frac{r^2}{x^3 r^2} = \frac{1}{x^3}$$ Step 3: Sum left side: $$\frac{1}{x^3 r^6} + \frac{1}{x^3 r^3} + \frac{1}{x^3} = \frac{1}{x^3} \left( \frac{1}{r^6} + \frac{1}{r^3} + 1 \right)$$ Step 4: Right side: $$\frac{1}{x^3} + \frac{1}{(xr)^3} + \frac{1}{(xr^2)^3} = \frac{1}{x^3} + \frac{1}{x^3 r^3} + \frac{1}{x^3 r^6} = \frac{1}{x^3} \left(1 + \frac{1}{r^3} + \frac{1}{r^6} \right)$$ Step 5: Both sides are equal, hence proved. --- 3. **Problem 8:** Given $x = \frac{2ab}{a+b}$, find value of $$\frac{x+a}{x-a} + \frac{x+b}{x-b}$$ Step 1: Substitute $x$: $$\frac{\frac{2ab}{a+b} + a}{\frac{2ab}{a+b} - a} + \frac{\frac{2ab}{a+b} + b}{\frac{2ab}{a+b} - b}$$ Step 2: Simplify numerator and denominator of first fraction: $$\frac{\frac{2ab + a(a+b)}{a+b}}{\frac{2ab - a(a+b)}{a+b}} = \frac{2ab + a^2 + ab}{2ab - a^2 - ab} = \frac{a^2 + 3ab}{ab - a^2}$$ Step 3: Factor denominator: $$ab - a^2 = a(b - a)$$ Step 4: Similarly for second fraction: $$\frac{\frac{2ab + b(a+b)}{a+b}}{\frac{2ab - b(a+b)}{a+b}} = \frac{2ab + ab + b^2}{2ab - ab - b^2} = \frac{3ab + b^2}{ab - b^2}$$ Denominator: $$ab - b^2 = b(a - b) = -b(b - a)$$ Step 5: Rewrite sum: $$\frac{a^2 + 3ab}{a(b - a)} + \frac{3ab + b^2}{-b(b - a)} = \frac{a^2 + 3ab}{a(b - a)} - \frac{3ab + b^2}{b(b - a)}$$ Step 6: Common denominator $ab(b - a)$: $$\frac{b(a^2 + 3ab) - a(3ab + b^2)}{ab(b - a)}$$ Step 7: Expand numerator: $$b a^2 + 3ab^2 - 3a^2 b - a b^2 = (b a^2 - 3a^2 b) + (3ab^2 - a b^2) = -2 a^2 b + 2 a b^2 = 2 a b (b - a)$$ Step 8: Substitute back: $$\frac{2 a b (b - a)}{a b (b - a)} = 2$$ Answer: $2$ --- 4. **Problem 9:** Given $$x = \frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}$$ Find value of $$\frac{x + 2\sqrt{2}}{x - 2\sqrt{2}} + \frac{x + 2\sqrt{3}}{x - 2\sqrt{3}}$$ Step 1: Rationalize denominator of $x$: $$x = \frac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{4\sqrt{6}(\sqrt{3} - \sqrt{2})}{3 - 2} = 4\sqrt{6}(\sqrt{3} - \sqrt{2})$$ Step 2: Simplify $x$: $$4\sqrt{6} \times \sqrt{3} = 4 \sqrt{18} = 4 \times 3 \sqrt{2} = 12 \sqrt{2}$$ $$4\sqrt{6} \times (-\sqrt{2}) = -4 \sqrt{12} = -4 \times 2 \sqrt{3} = -8 \sqrt{3}$$ So, $$x = 12 \sqrt{2} - 8 \sqrt{3}$$ Step 3: Calculate first fraction numerator and denominator: $$x + 2\sqrt{2} = 12 \sqrt{2} - 8 \sqrt{3} + 2 \sqrt{2} = 14 \sqrt{2} - 8 \sqrt{3}$$ $$x - 2\sqrt{2} = 12 \sqrt{2} - 8 \sqrt{3} - 2 \sqrt{2} = 10 \sqrt{2} - 8 \sqrt{3}$$ Step 4: Calculate second fraction numerator and denominator: $$x + 2\sqrt{3} = 12 \sqrt{2} - 8 \sqrt{3} + 2 \sqrt{3} = 12 \sqrt{2} - 6 \sqrt{3}$$ $$x - 2\sqrt{3} = 12 \sqrt{2} - 8 \sqrt{3} - 2 \sqrt{3} = 12 \sqrt{2} - 10 \sqrt{3}$$ Step 5: Simplify each fraction by rationalizing denominators: First fraction: Multiply numerator and denominator by conjugate of denominator: $$(10 \sqrt{2} + 8 \sqrt{3})$$ Numerator: $$(14 \sqrt{2} - 8 \sqrt{3})(10 \sqrt{2} + 8 \sqrt{3}) = 14 \sqrt{2} \times 10 \sqrt{2} + 14 \sqrt{2} \times 8 \sqrt{3} - 8 \sqrt{3} \times 10 \sqrt{2} - 8 \sqrt{3} \times 8 \sqrt{3}$$ $$= 140 \times 2 + 112 \sqrt{6} - 80 \sqrt{6} - 64 \times 3 = 280 + 32 \sqrt{6} - 192 = 88 + 32 \sqrt{6}$$ Denominator: $$(10 \sqrt{2} - 8 \sqrt{3})(10 \sqrt{2} + 8 \sqrt{3}) = (10 \sqrt{2})^2 - (8 \sqrt{3})^2 = 100 \times 2 - 64 \times 3 = 200 - 192 = 8$$ First fraction simplified: $$\frac{88 + 32 \sqrt{6}}{8} = 11 + 4 \sqrt{6}$$ Step 6: Second fraction: Multiply numerator and denominator by conjugate of denominator: $$(12 \sqrt{2} + 10 \sqrt{3})$$ Numerator: $$(12 \sqrt{2} - 6 \sqrt{3})(12 \sqrt{2} + 10 \sqrt{3}) = 12 \sqrt{2} \times 12 \sqrt{2} + 12 \sqrt{2} \times 10 \sqrt{3} - 6 \sqrt{3} \times 12 \sqrt{2} - 6 \sqrt{3} \times 10 \sqrt{3}$$ $$= 144 \times 2 + 120 \sqrt{6} - 72 \sqrt{6} - 60 \times 3 = 288 + 48 \sqrt{6} - 180 = 108 + 48 \sqrt{6}$$ Denominator: $$(12 \sqrt{2} - 10 \sqrt{3})(12 \sqrt{2} + 10 \sqrt{3}) = (12 \sqrt{2})^2 - (10 \sqrt{3})^2 = 144 \times 2 - 100 \times 3 = 288 - 300 = -12$$ Second fraction simplified: $$\frac{108 + 48 \sqrt{6}}{-12} = -9 - 4 \sqrt{6}$$ Step 7: Sum both fractions: $$(11 + 4 \sqrt{6}) + (-9 - 4 \sqrt{6}) = 2$$ Answer: $2$ --- 5. **Problem 10:** Solve $$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$$ Step 1: Cross multiply: $$91(x^3 + 3x) = 341(3x^2 + 1)$$ Step 2: Expand: $$91x^3 + 273x = 1023x^2 + 341$$ Step 3: Rearrange: $$91x^3 - 1023x^2 + 273x - 341 = 0$$ Step 4: Try to factor or find rational roots using Rational Root Theorem. Possible roots are factors of 341 over factors of 91. Try $x=7$: $$91(343) - 1023(49) + 273(7) - 341$$ Calculate: $$91 \times 343 = 31213$$ $$1023 \times 49 = 50127$$ $$273 \times 7 = 1911$$ Sum: $$31213 - 50127 + 1911 - 341 = 31213 - 50127 + 1570 = -18914 + 1570 = -17344 \neq 0$$ Try $x= 7/13$ (since 13 divides 91 and 341): Calculate each term: $$x = \frac{7}{13}$$ $$x^3 = \frac{343}{2197}$$ $$x^2 = \frac{49}{169}$$ Calculate left side numerator: $$x^3 + 3x = \frac{343}{2197} + 3 \times \frac{7}{13} = \frac{343}{2197} + \frac{21}{13}$$ Find common denominator $2197 = 13^3$: $$\frac{343}{2197} + \frac{21 \times 169}{2197} = \frac{343 + 3549}{2197} = \frac{3892}{2197}$$ Calculate denominator: $$3x^2 + 1 = 3 \times \frac{49}{169} + 1 = \frac{147}{169} + 1 = \frac{147 + 169}{169} = \frac{316}{169}$$ Fraction: $$\frac{3892/2197}{316/169} = \frac{3892}{2197} \times \frac{169}{316} = \frac{3892 \times 169}{2197 \times 316}$$ Calculate numerator: $$3892 \times 169 = 3892 \times (170 - 1) = 3892 \times 170 - 3892 = 661640 - 3892 = 657748$$ Calculate denominator: $$2197 \times 316 = 2197 \times (300 + 16) = 659100 + 35152 = 694252$$ Fraction: $$\frac{657748}{694252} \approx 0.947$$ Right side: $$\frac{341}{91} \approx 3.747$$ Not equal, so $x=7/13$ is not root. Step 5: Use substitution or numerical methods. Alternatively, note the equation is of the form: $$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$$ Rewrite as: $$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$$ Try to express as a continued proportion or use the property of proportion: $$\frac{x^3 + 3x}{3x^2 + 1} = \frac{a}{b}$$ Try $x = 7$: $$\frac{343 + 21}{147 + 1} = \frac{364}{148} = \frac{91}{37} \neq \frac{341}{91}$$ Try $x = 3$: $$\frac{27 + 9}{27 + 1} = \frac{36}{28} = \frac{9}{7} \neq \frac{341}{91}$$ Try $x = 1$: $$\frac{1 + 3}{3 + 1} = \frac{4}{4} = 1 \neq \frac{341}{91}$$ Try $x = 13/7$: Calculate numerator: $$\left(\frac{13}{7}\right)^3 + 3 \times \frac{13}{7} = \frac{2197}{343} + \frac{39}{7} = \frac{2197}{343} + \frac{1911}{343} = \frac{4108}{343}$$ Denominator: $$3 \times \left(\frac{13}{7}\right)^2 + 1 = 3 \times \frac{169}{49} + 1 = \frac{507}{49} + 1 = \frac{556}{49}$$ Fraction: $$\frac{4108/343}{556/49} = \frac{4108}{343} \times \frac{49}{556} = \frac{4108 \times 49}{343 \times 556}$$ Calculate numerator: $$4108 \times 49 = 201292$$ Calculate denominator: $$343 \times 556 = 190708$$ Fraction: $$\frac{201292}{190708} \approx 1.055 \neq \frac{341}{91} \approx 3.747$$ Step 6: Since direct substitution is complicated, solve cubic equation numerically or by factorization. Step 7: Alternatively, multiply both sides by denominator and rearrange: $$91x^3 + 273x = 1023x^2 + 341$$ $$91x^3 - 1023x^2 + 273x - 341 = 0$$ Try to factor by grouping or use rational root theorem. Try $x=7$ again: $$91(343) - 1023(49) + 273(7) - 341 = 31213 - 50127 + 1911 - 341 = -1344 \neq 0$$ Try $x= 13$: $$91(2197) - 1023(169) + 273(13) - 341 = 199927 - 172887 + 3549 - 341 = 300248 \neq 0$$ Try $x= 1$: $$91 - 1023 + 273 - 341 = 0$$ Calculate: $$91 - 1023 = -932$$ $$-932 + 273 = -659$$ $$-659 - 341 = -1000 \neq 0$$ Try $x= 11$: $$91(1331) - 1023(121) + 273(11) - 341 = 121121 - 123783 + 3003 - 341 = 0$$ Calculate: $$121121 - 123783 = -2662$$ $$-2662 + 3003 = 341$$ $$341 - 341 = 0$$ So $x=11$ is a root. Step 8: Divide polynomial by $(x - 11)$: Coefficients: 91, -1023, 273, -341 Bring down 91. Multiply 91 by 11 = 1001; add to -1023 = -22. Multiply -22 by 11 = -242; add to 273 = 31. Multiply 31 by 11 = 341; add to -341 = 0. Quotient polynomial: $$91x^2 - 22x + 31$$ Step 9: Solve quadratic: $$91x^2 - 22x + 31 = 0$$ Discriminant: $$\Delta = (-22)^2 - 4 \times 91 \times 31 = 484 - 11284 = -10800 < 0$$ No real roots. Step 10: Final solution: $$x = 11$$ --- **Final answers:** 6. $k = 4$, factorisation: $(x - 1)(3x - 1)(x + 4)$ 7. Proven identity holds. 8. Value is $2$ 9. Value is $2$ 10. $x = 11$