39) \( h a d x \rightarrow 3 \frac{x^{3}-5 x^{2}+6 x}{x^{2}-3 x} \)

To simplify the expression \( h a d x \rightarrow 3 \frac{x^{3}-5 x^{2}+6 x}{x^{2}-3 x} \), start by factoring both the numerator and the denominator. The numerator \( x^{3}-5x^{2}+6x \) can be factored as \( x(x^{2}-5x+6) = x(x-2)(x-3) \). The denominator \( x^{2}-3x \) can be factored as \( x(x-3) \). Now, the expression simplifies to: \[ 3 \frac{x(x-2)(x-3)}{x(x-3)} = 3(x-2) \quad \text{(as long as \( x \neq 0 \) and \( x \neq 3 \))} \] So, the simplified form is \( h a d x \rightarrow 3(x-2) \). To ensure you're on the right track with algebraic simplifications, remember that canceling terms can make mistakes. A common mishap is overlooking restrictions. When canceling \( x \) and \( x-3 \), always highlight the values that make the original function undefined. In this case, \( x \neq 0 \) and \( x \neq 3 \) to prevent division by zero. Keeping track of these restrictions is crucial in calculus and algebra!