E/P) 17 Find the remainder when \( x^{3}+6 x^{2}+5 x-12 \) is divided by \( i x-2 \), ii \( x+3 \). b Hence, or otherwise, find all the solutions to the equation \( x^{3}+6 x^{2}+5 x-12=0 \)

Let's solve the problem step by step. ### Given Polynomial: \[ f(x) = x^{3} + 6x^{2} + 5x - 12 \] ### Part (a): #### (i) Remainder when \( f(x) \) is divided by \( ix - 2 \): To find the remainder when \( f(x) \) is divided by \( ix - 2 \), we can use the **Remainder Theorem**. The theorem states that the remainder of a polynomial \( f(x) \) divided by \( x - a \) is \( f(a) \). First, solve for \( x \) in the divisor \( ix - 2 = 0 \): \[ ix - 2 = 0 \implies x = \frac{2}{i} \] Recall that \( \frac{1}{i} = -i \) (since \( i \times (-i) = 1 \)). Therefore: \[ x = \frac{2}{i} = -2i \] Now, evaluate \( f(x) \) at \( x = -2i \): \[ f(-2i) = (-2i)^3 + 6(-2i)^2 + 5(-2i) - 12 \] Calculate each term: \[ (-2i)^3 = -8i^3 = -8(-i) = 8i \\ 6(-2i)^2 = 6 \times 4i^2 = 6 \times 4(-1) = -24 \\ 5(-2i) = -10i \\ \] Combine the terms: \[ f(-2i) = 8i - 24 - 10i - 12 = (-2i) - 36 = -36 - 2i \] **Remainder:** \(-36 - 2i\) #### (ii) Remainder when \( f(x) \) is divided by \( x + 3 \): Using the Remainder Theorem again, set \( x + 3 = 0 \), so \( x = -3 \). Evaluate \( f(-3) \): \[ f(-3) = (-3)^3 + 6(-3)^2 + 5(-3) - 12 = -27 + 54 - 15 - 12 = 0 \] **Remainder:** \(0\) ### Part (b): Finding All Solutions to \( f(x) = 0 \) Since the remainder is zero when \( f(x) \) is divided by \( x + 3 \), \( x = -3 \) is a root of the polynomial. We can factor \( f(x) \) as follows: \[ f(x) = (x + 3)(x^{2} + 3x - 4) \] Now, solve the quadratic equation \( x^{2} + 3x - 4 = 0 \): \[ x = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \] This gives two solutions: \[ x = \frac{-3 + 5}{2} = 1 \quad \text{and} \quad x = \frac{-3 - 5}{2} = -4 \] ### **All Solutions:** \[ x = -3, \quad x = 1, \quad x = -4 \] ### **Final Answer:** **a.** 1. The remainder when divided by \( ix - 2 \) is \( -36 - 2i \). 2. The remainder when divided by \( x + 3 \) is \( 0 \). **b.** All solutions to the equation \( x^{3} + 6x^{2} + 5x - 12 = 0 \) are: \[ x = -3,\quad x = 1,\quad \text{and}\quad x = -4. \]