13. Let \( f(x)=x^{3}-5 x^{2}+2 x+8 \) be a polynomial. (i) Show that \( x+1 \) is a factor of the polynomial \( f(x) \) (ii) Factorize the polynomial \( f(x) \) completely. (iii) Calculate the range of values of \( x \) for which \( 2 x+8>5 x^{2}-x^{3} \)

To tackle this polynomial problem, let's break it down into simpler steps. (i) To show that \( x+1 \) is a factor of the polynomial \( f(x) \), we can use the Remainder Theorem, which states that if a polynomial \( f(x) \) is divided by \( x-c \) and the remainder is zero, then \( x-c \) is a factor of \( f(x) \). Here, we will substitute \( x = -1 \) into \( f(x) \): \[ f(-1) = (-1)^3 - 5(-1)^2 + 2(-1) + 8 = -1 - 5 - 2 + 8 = 0. \] Since \( f(-1) = 0 \), it follows that \( x+1 \) is a factor of \( f(x) \). (ii) To factorize \( f(x) \) completely, we will perform synthetic division of \( f(x) \) by \( x+1 \): ``` -1 | 1 -5 2 8 | -1 6 -8 -------------------- 1 -6 8 0 ``` The result yields \( f(x) = (x + 1)(x^2 - 6x + 8) \). We can further factor \( x^2 - 6x + 8 \): \[ x^2 - 6x + 8 = (x - 2)(x - 4). \] Thus, the complete factorization of \( f(x) \) is: \[ f(x) = (x + 1)(x - 2)(x - 4). \] (iii) Now, to solve the inequality \( 2x + 8 > 5x^2 - x^3 \), we can rearrange it: \[ x^3 - 5x^2 + 2x + 8 > 0. \] We know the roots from our previous factorization \( f(x) \). The roots are \( x = -1, 2, 4 \). Testing intervals determined by these roots: 1. For \( x < -1 \), say \( x = -2 \): \( (-2 + 1)(-2 - 2)(-2 - 4) = (-1)(-4)(-6) < 0 \). 2. For \( -1 < x < 2 \), say \( x = 0 \): \( (0 + 1)(0 - 2)(0 - 4) = (1)(-2)(-4) > 0 \). 3. For \( 2 < x < 4 \), say \( x = 3 \): \( (3 + 1)(3 - 2)(3 - 4) = (4)(1)(-1) < 0 \). 4. For \( x > 4 \), say \( x = 5 \): \( (5 + 1)(5 - 2)(5 - 4) = (6)(3)(1) > 0 \). Thus, the ranges where \( f(x) > 0 \) are \( (-1, 2) \) and \( (4, \infty) \).