(b) f - g
To solve it, we have to do f(x) - g(x).
[tex]\begin{gathered} f(x)-g(x) \\ 2x^2-14x+20-(x-5) \\ 2x^2-14x+20-x+5 \\ 2x^2-15x+25 \end{gathered}[/tex]And, since there are no restrictions for the domains,
The domain is (-∞,∞).
(c) f * g
To solve it, we have to multiplicate both functions.
[tex]\begin{gathered} f(x)\cdot g(x) \\ 2x^2-14x+20\cdot(x-5) \\ 2x^3-14x^2+20x-10x^2+70x-100 \\ 2x^3-24x^2+90x-100 \end{gathered}[/tex]And, since there are no restrictions for the domain,
The domain is (-∞,∞).
(d) f / g
To solve it, we have to divide f(x) by g(x).
To divide it, an easy way is to factor f(x).
So, let's find the zeros of the function using Bhaskara.
[tex]\begin{gathered} y=2x^2-14x+20 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{+14\pm\sqrt[]{14^2-4\cdot2\cdot20}}{2\cdot2} \\ x=\frac{+14\pm\sqrt[]{196-160}}{4} \\ x=\frac{+14\pm\sqrt[]{36}}{4}=\frac{+14\pm6}{4} \\ x_1=\frac{+14+6}{4}=\frac{20}{4}=5 \\ x_2=\frac{+14-6}{4}=\frac{8}{4}=2 \end{gathered}[/tex]So, since 5 and 2 are the zeros of the function, and the coefficient before the quadratic term is 2, f(x) can be written as:
[tex]f(x)=2\cdot(x-5)\cdot(x-2)[/tex]So, now we can divide f(x) by g(x).
[tex]\begin{gathered} \frac{f(x)}{g(x)} \\ \frac{2x^2-14x+20}{x-5}=\frac{2\cdot(x-5)\cdot(x-2)}{(x-5)} \\ =2\cdot(x-2) \end{gathered}[/tex]And, since there are no restrictions for the domain,
The domain is (-∞,∞).