1. The problem involves finding values of $a$ and $b$ for the polynomial $f(x) = ax^3 - 15x^2 + bx - 2$ where $2x - 1$ is a factor and the remainder when divided by $x-1$ is 5. 2. Since $2x - 1$ is a factor, $f(\frac{1}{2}) = 0$. Calculate $f(\frac{1}{2}) = a(\frac{1}{2})^3 - 15(\frac{1}{2})^2 + b(\frac{1}{2}) - 2 = 0$. This simplifies to: $$ \frac{a}{8} - \frac{15}{4} + \frac{b}{2} - 2 = 0 $$ Multiply through by 8 to clear denominators: $$ a - 30 + 4b - 16 = 0 $$ $$ a + 4b = 46 \quad (1) $$ 3. The remainder when divided by $x-1$ is 5, so $f(1) = 5$. Calculate $f(1) = a(1)^3 - 15(1)^2 + b(1) - 2 = 5$. Simplify: $$ a - 15 + b - 2 = 5 $$ $$ a + b = 22 \quad (2) $$ 4. From equations (1) and (2): (1) $a + 4b = 46$ (2) $a + b = 22$ Subtract (2) from (1): $$ (a + 4b) - (a + b) = 46 - 22 $$ $$ 3b = 24 $$ $$ b = 8 $$ Substitute $b = 8$ into (2): $$ a + 8 = 22 $$ $$ a = 14 $$ 5. Using $a = 14$ and $b = 8$, express $f(x)$ as $(2x - 1)g(x)$, where $g(x)$ is quadratic. Perform polynomial division of $f(x) = 14x^3 -15x^2 + 8x - 2$ by $2x - 1$. Divide leading terms: $14x^3 \div 2x = 7x^2$. Multiply divisor by $7x^2$: $14x^3 - 7x^2$. Subtract: $(-15x^2) - (-7x^2) = -8x^2$. Bring down $8x$. Divide $-8x^2 \div 2x = -4x$. Multiply divisor by $-4x$: $-8x^2 + 4x$. Subtract: $(8x) - (4x) = 4x$. Bring down $-2$. Divide $4x \div 2x = 2$. Multiply divisor by $2$: $4x - 2$. Subtract: $(-2) - (-2) = 0$. Therefore, $g(x) = 7x^2 - 4x + 2$. Final expression: $$ f(x) = (2x - 1)(7x^2 - 4x + 2) $$ 2. Given $f(x) = 6x^3 - 5x^2 + ax + b$ with factor $x+2$ and remainder 27 when divided by $x-1$. (i) Since $x + 2$ is a factor, then $f(-2) = 0$: $$ f(-2) = 6(-2)^3 - 5(-2)^2 + a(-2) + b = 0$$ $$ -48 - 20 - 2a + b = 0 $$ $$ -68 - 2a + b = 0 \\ b = 68 + 2a \quad (3) $$ (ii) Remainder when divided by $x - 1$ is 27, so $f(1) = 27$: $$ 6 - 5 + a + b = 27 $$ $$ 1 + a + b = 27 $$ $$ a + b = 26 \quad (4) $$ (iii) Substitute (3) into (4): $$ a + 68 + 2a = 26 $$ $$ 3a + 68 = 26 $$ $$ 3a = -42 $$ $$ a = -14 $$ Substitute $a$ back to find $b$: $$ b = 68 + 2(-14) = 68 - 28 = 40 $$ 6. Using $a = -14$ and $b = 40$, Express $f(x)$ as $(x + 2)(px^2 + qx + r)$. Divide $f(x) = 6x^3 - 5x^2 - 14x + 40$ by $x + 2$. Leading term: $6x^3 \div x = 6x^2$. Multiply divisor: $6x^3 + 12x^2$. Subtract: $-5x^2 - 12x^2 = -17x^2$. Bring down $-14x$. Divide: $-17x^2 \div x = -17x$. Multiply divisor: $-17x^2 -34x$. Subtract: $-14x - (-34x) = 20x$. Bring down $40$. Divide $20x \div x = 20$. Multiply divisor: $20x + 40$. Subtract: $40 - 40 = 0$. So, $p = 6$, $q = -17$, $r = 20$ and $$ f(x) = (x + 2)(6x^2 - 17x + 20) $$ 7. Solve $f(x) = 0$: Set each factor to zero: $$ x + 2 = 0 \implies x = -2 $$ Solve quadratic $6x^2 - 17x + 20 = 0$. Calculate discriminant: $$ \Delta = (-17)^2 - 4 \times 6 \times 20 = 289 - 480 = -191 < 0 $$ No real roots from quadratic factor. Therefore, the only real root is $x = -2$. 3. Polynomial $p(x) = ax^3 - 9x^2 + bx - 6$, factor $x - 2$, remainder 66 when divided by $x - 3$. (a) Since $x - 2$ is factor, $p(2) = 0$: $$ a(2)^3 - 9(2)^2 + b(2) - 6 = 0 $$ $$ 8a - 36 + 2b - 6 = 0 $$ $$ 8a + 2b = 42 \quad (5) $$ Remainder when divided by $x - 3$ is 66, so $p(3) = 66$: $$ 27a - 81 + 3b - 6 = 66 $$ $$ 27a + 3b = 153 \quad (6) $$ Divide (6) by 3: $$ 9a + b = 51 \quad (7) $$ From (5), divide by 2: $$ 4a + b = 21 \quad (8) $$ Subtract (8) from (7): $$ (9a + b) - (4a + b) = 51 - 21 $$ $$ 5a = 30 \implies a = 6 $$ Substitute into (8): $$ 4(6) + b = 21 \implies 24 + b = 21 \implies b = -3 $$ (b) Now express $p(x)$ as $(x - 2)q(x)$. Divide $6x^3 - 9x^2 - 3x - 6$ by $x - 2$. Divide leading terms: $6x^3 \div x = 6x^2$. Multiply divisor: $6x^3 - 12x^2$. Subtract: $-9x^2 - (-12x^2) = 3x^2$. Bring down $-3x$. Divide $3x^2 \div x = 3x$. Multiply divisor: $3x^2 - 6x$. Subtract: $-3x - (-6x) = 3x$. Bring down $-6$. Divide $3x \div x = 3$. Multiply divisor: $3x - 6$. Subtract: $-6 - (-6) = 0$. Then $$ q(x) = 6x^2 + 3x + 3 $$ (c) To check the number of real solutions solve $p(x) = 0$: Because $p(x) = (x - 2)(6x^2 + 3x + 3)$. Calculate discriminant of quadratic: $$ \Delta = (3)^2 - 4 \times 6 \times 3 = 9 - 72 = -63 < 0 $$ No real roots for quadratic. Hence, only one real solution to $p(x) = 0$: $$ x = 2 $$ 4. Polynomial $f(x) = 4x^3 + kx + p$ is divisible by $x + 2$ and its derivative $f'(x)$ divisible by $2x - 1$. (i) Since divisible by $x + 2$, $f(-2) = 0$: $$ f(-2) = 4(-2)^3 + k(-2) + p = -32 - 2k + p = 0 $$ Derivative: $$ f'(x) = 12x^2 + k $$ Since divisible by $2x - 1$, then $f'(\frac{1}{2}) = 0$: $$ f'\left(\frac{1}{2}\right) = 12 \left(\frac{1}{2}\right)^2 + k = 12 \times \frac{1}{4} + k = 3 + k = 0 $$ $$ k = -3 $$ Substitute $k = -3$ into $f(-2) = 0$: $$ -32 - 2(-3) + p = 0 \implies -32 + 6 + p = 0 \implies p = 26 $$ (ii) Using $k = -3$ and $p = 26$, express $f(x)$ as $(x + 2)(ax^2 + bx + c)$. Divide: $$ 4x^3 - 3x + 26 $$ by $(x + 2)$. Leading term: $$4x^3 \div x = 4x^2$$ Multiply divisor: $$4x^3 + 8x^2$$ Subtract: $$-3x + 26 - 8x^2 = -8x^2 - 3x + 26$$ Divide: $$-8x^2 \div x = -8x$$ Multiply divisor: $$-8x^2 - 16x$$ Subtract: $$(-3x) - (-16x) = 13x$$ Bring down 26. Divide: $$13x \div x = 13$$ Multiply divisor: $$13x + 26$$ Subtract remainder: $$26 - 26 = 0$$ Thus, $$ f(x) = (x + 2)(4x^2 - 8x + 13) $$ (iii) To find solutions of $f(x) = 0$, we solve: $$ (x + 2)(4x^2 - 8x + 13) = 0 $$ First factor: $$ x = -2 $$ Second factor: $$ 4x^2 - 8x + 13 = 0 $$ Calculate discriminant: $$ \Delta = (-8)^2 - 4 \times 4 \times 13 = 64 - 208 = -144 < 0 $$ No real roots from quadratic factor. Hence, only one real solution to $f(x) = 0$ is: $$ x = -2 $$