Let f(x)=x^2+9x+20 and g(x)=x^3-4x Find (fg)(x) and ((f)/(g))(x) (fg)(x)=vert x^5+5x^4-16x^2-80x ((f)/(g))(x)=vert (x+4)/(x-2)x State the domain of each. Domain of (fg)(x) all real numbers Domain of ((f)/(g))(x) xvert xneq 0,2 : all real numbers xvert xneq 0 xvert xneq -2,0,2 xvert xneq 0,2 Evaluate the following. (fg)(-1)=66x ((f)/(g))(-1)=vert -1x

To solve the given problem, we need to find \((fg)(x)\) and \(\left(\frac{f}{g}\right)(x)\), and then determine their domains.1. **Finding \((fg)(x)\):** Given: The product \((fg)(x)\) is: Expanding this product: So, \((fg)(x) = x^5 + 9x^4 + 16x^3 - 36x^2 - 80x\).2. **Finding \(\left(\frac{f}{g}\right)(x)\):** The quotient \(\left(\frac{f}{g}\right)(x)\) is: To simplify, factor both the numerator and the denominator: - Numerator: \(x^2 + 9x + 20 = (x + 4)(x + 5)\) - Denominator: \(x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)\) Thus: 3. **Domain of \((fg)(x)\):** Since \((fg)(x)\) is a polynomial, its domain is all real numbers.4. **Domain of \(\left(\frac{f}{g}\right)(x)\):** The domain excludes values that make the denominator zero: Therefore, the domain is .5. **Evaluate \((fg)(-1)\):** Substitute into \((fg)(x)\): 6. **Evaluate \(\left(\frac{f}{g}\right)(-1)\):** Substitute into \(\left(\frac{f}{g}\right)(x)\): In summary:- \((fg)(x) = x^5 + 9x^4 + 16x^3 - 36x^2 - 80x\)- Domain of \((fg)(x)\): all real numbers- \(\left(\frac{f}{g}\right)(x) = \frac{(x + 4)(x + 5)}{x(x - 2)(x + 2)}\)- Domain of \(\left(\frac{f}{g}\right)(x)\): - \((fg)(-1) = 36\)- \(\left(\frac{f}{g}\right)(-1) = 4\)