The function \( f \) is defined as \( f(x)=\frac{6}{-5 x^{2}+3 x} \) Find \( f(x+1) \). Write your answer without parentheses, and simplify it as much as possible. \[ (x+1)=\square \]

We start with \[ f(x) = \frac{6}{-5x^2 + 3x}. \] Substitute \(x+1\) for \(x\) to obtain \[ f(x+1) = \frac{6}{-5(x+1)^2 + 3(x+1)}. \] First, expand the squared term: \[ (x+1)^2 = x^2 + 2x + 1. \] Thus, \[ -5(x+1)^2 = -5x^2 - 10x - 5. \] Now, distribute the \(3\) in the second term: \[ 3(x+1) = 3x + 3. \] Combine these results for the denominator: \[ -5x^2 - 10x - 5 + 3x + 3 = -5x^2 - 7x - 2. \] So we have: \[ f(x+1) = \frac{6}{-5x^2 - 7x - 2}. \] Next, factor the denominator. Notice that multiplying the entire expression by \(-1\) gives: \[ -5x^2 - 7x - 2 = -\left(5x^2 + 7x + 2\right). \] Factor the quadratic \(5x^2 + 7x + 2\). We look for two numbers that multiply to \(5 \cdot 2 = 10\) and add to \(7\); these are \(5\) and \(2\). Hence, \[ 5x^2 + 7x + 2 = (5x + 2)(x + 1). \] Thus, the denominator becomes: \[ -5x^2 - 7x - 2 = - (5x+2)(x+1). \] Substitute back into the function: \[ f(x+1)=\frac{6}{- (5x+2)(x+1)}. \] This simplifies to: \[ f(x+1)=-\frac{6}{(5x+2)(x+1)}. \] Since we need the answer without any extra parentheses, we write the product in the denominator as a polynomial by multiplying out: \[ (5x+2)(x+1)=5x^2+7x+2. \] Thus, the simplified answer is: \[ -\frac{6}{5x^2+7x+2}. \] \[ (x+1)=-\frac{6}{5x^2+7x+2} \]