1. **Resolve into partial fractions:** (i) $$\frac{6x - 10}{x^2 - 2x - 3}$$ Factor denominator: $$x^2 - 2x - 3 = (x - 3)(x + 1)$$ Set $$\frac{6x - 10}{(x - 3)(x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1}$$ Multiply both sides by denominator: $$6x - 10 = A(x + 1) + B(x - 3)$$ Expand: $$6x - 10 = A x + A + B x - 3B = (A + B)x + (A - 3B)$$ Equate coefficients: $$6 = A + B$$ $$-10 = A - 3B$$ Solve system: From first, $$A = 6 - B$$ Substitute into second: $$-10 = (6 - B) - 3B = 6 - 4B$$ $$-16 = -4B \Rightarrow B = 4$$ Then $$A = 6 - 4 = 2$$ Answer: $$\frac{6x - 10}{x^2 - 2x - 3} = \frac{2}{x - 3} + \frac{4}{x + 1}$$ --- 2. **Find remainders when dividing polynomials:** Use Remainder Theorem: remainder of $$f(x)$$ divided by $$x - c$$ is $$f(c)$$. (i)(a) Divide $$x^2 - 5x + 1$$ by $$x - 2$$: $$f(2) = 2^2 - 5(2) + 1 = 4 - 10 + 1 = -5$$ (i)(b) Divide by $$x + 1$$ (i.e., $$x - (-1)$$): $$f(-1) = (-1)^2 - 5(-1) + 1 = 1 + 5 + 1 = 7$$ --- 3. **Find value of k if (x - 1) is a factor of $$x^3 + x^2 + kx + 1$$:** Since (x - 1) is a factor, remainder when dividing by (x - 1) is zero: $$f(1) = 1^3 + 1^2 + k(1) + 1 = 1 + 1 + k + 1 = 3 + k = 0$$ Solve: $$k = -3$$ --- 4. **(x - 1) and (x + 3) are factors of $$x^3 + ax^2 + bx + 12$$. Find a, b, and remainder:** Since both are factors, $$f(1) = 0$$ and $$f(-3) = 0$$. Calculate: $$f(1) = 1 + a + b + 12 = 0 \Rightarrow a + b = -13$$ $$f(-3) = (-3)^3 + a(-3)^2 + b(-3) + 12 = -27 + 9a - 3b + 12 = 0$$ Simplify: $$-15 + 9a - 3b = 0 \Rightarrow 9a - 3b = 15$$ Divide by 3: $$3a - b = 5$$ Solve system: From first, $$b = -13 - a$$ Substitute into second: $$3a - (-13 - a) = 5 \Rightarrow 3a + 13 + a = 5 \Rightarrow 4a = -8 \Rightarrow a = -2$$ Then $$b = -13 - (-2) = -11$$ Remainder of $$f(x) = 0$$ since both are factors. --- 5. **Find zeros of $$x^3 + 2x^2 - 5x - 6$$:** Try rational roots: factors of 6 are ±1, ±2, ±3, ±6. Test $$x=1$$: $$1 + 2 - 5 - 6 = -8 \neq 0$$ Test $$x=2$$: $$8 + 8 - 10 - 6 = 0$$ So $$x=2$$ is a root. Divide polynomial by $$x - 2$$: Quotient: $$x^2 + 4x + 3$$ Factor quadratic: $$(x + 1)(x + 3)$$ Zeros are $$x = 2, -1, -3$$ --- 6. **Show $$x - a$$ is a factor of $$x^3 - a^3$$ and find other factor:** Use difference of cubes formula: $$x^3 - a^3 = (x - a)(x^2 + ax + a^2)$$ Hence factorization is: $$(x - a)(x^2 + ax + a^2)$$ --- 7. **When $$a + 7x + bx^2$$ is divided by $$x + 2$$, remainder is -1:** Evaluate at $$x = -2$$: $$a + 7(-2) + b(-2)^2 = a - 14 + 4b = -1$$ Rearranged: $$a + 4b = 13$$ --- **Summary:** - Partial fractions example shown for (i). - Remainders found using substitution. - Factor conditions used to find unknown coefficients. - Polynomial roots found by rational root test and division. - Factorization of difference of cubes applied. - Remainder condition used to relate coefficients. q_count: 7