Question

reduce the rational expression to lowest terms.\\(\frac{x^{2}-3x - 40}{x^{2}-8x + 8}\\)\\(\frac{x^{2}-3x - 40}{x^{2}-8x + 8}=\square\\)

Explanation:

Step1: Factor the numerator

Factor \(x^{2}-3x - 40\). We need two numbers that multiply to - 40 and add up to - 3. The numbers are - 8 and 5. So \(x^{2}-3x - 40=(x - 8)(x+5)\).

Step2: Factor the denominator

Factor \(x^{2}-8x + 8\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c\), here \(a = 1\), \(b=-8\), \(c = 8\). But we can also try to factor by finding two numbers that multiply to 8 and add up to - 8. Since it doesn't factor nicely over the integers, we assume there is a mistake and it might be \(x^{2}-8x + 16=(x - 4)^{2}\). Let's work with the correct - factored form if it was a mis - type. If we assume the denominator is \(x^{2}-8x + 16=(x - 4)^{2}\), the rational expression \(\frac{x^{2}-3x - 40}{x^{2}-8x + 16}=\frac{(x - 8)(x + 5)}{(x - 4)^{2}}\) and it is already in lowest terms as there are no common factors in the numerator and the denominator.

Answer:

\(\frac{(x - 8)(x + 5)}{x^{2}-8x + 8}\) (if the denominator is as given) or \(\frac{(x - 8)(x + 5)}{(x - 4)^{2}}\) (if the denominator was meant to be \(x^{2}-8x + 16\))