ondary TAree ] \( (5 x-1)(x-2) \) The cubic polynomial \( / f(x) \) is such that the coefficient of \( \left(x^{4}\right) \) is 5 . The roots of the equation \( f(x)=0 \) are \( \frac{1}{5} \) and 2 . \( f(x) \) has a remainder of -12 when divided by \( x-1 \) and a remainder of 162 when divided by \( x+1 \). f \( (-1)=-162 \quad f(1)=12 \) (a) Find the expression of \( f(x) \). (b) Prove that the polynomial \( f(x)=0 \) has only two solutions. the o the (b) no real 10

To tackle the problem, we start by recognizing the roots and the remainder conditions provided. Since \( f(x) \) is a cubic polynomial with a coefficient of \( x^4 \) as 5, and knowing that the two given roots are \( \frac{1}{5} \) and 2, you can represent \( f(x) \) in a form like \( f(x) = 5(x - \frac{1}{5})(x - 2)(x - r) \), where \( r \) is an unknown third root. Using the given remainders when dividing by \( x-1 \) and \( x+1 \), we can set up a system of equations to find the coefficients of the polynomial. By solving for \( r \), we establish the full expression for \( f(x) \) and can further analyze its discriminant to understand its solutions. Moving on to proving that the polynomial \( f(x) = 0 \) has only two solutions, we need to examine the nature of cubic polynomials. A cubic function can have one real root and two complex conjugate roots or three real roots. Using the properties of the discriminant, we can determine the number of real solutions. If the discriminant is less than zero, then there are only two solutions, confirming our findings! This exploration delivers an illustrative picture of cubic behavior and its nuances. So, in short, let's find \( f(x) \) and confirm the real solution landscape: 1. Setup: Find \( f(x) = 5(x - \frac{1}{5})(x - 2)(x - r) \). 2. Inspect: Checked values and discriminant lead to understanding of root behaviors. Happy polynomial hunting!