Question 74672
1.
You're simply going to add each like term from both equations
{{{(f+g)(x)=f(x)+g(x)=(2x^2+5x+7)+(x^3+2)}}}
{{{(f+g)(x)=f(x)+g(x)=x^3+2x^2+5x+9}}}
Domain is all real numbers since the domain of the 2 equations is all real numbers. 




2.
Now you're going to subtract like terms.
{{{(f-g)(x)=f(x)-g(x)=(2x^2+5x+7)-(x^3+2)}}}
{{{(f-g)(x)=f(x)-g(x)=-x^3+2x^2+5x+5}}}
Domain is all real numbers since the domain of the 2 equations is all real numbers. 



3. 
Now you're going to multiply each equation
{{{(f*g)(x)=f(x)*g(x)=(2x^2+5x+7)(x^3+2)}}}
{{{(f*g)(x)=f(x)*g(x)=2x^2*x^3+2x^2(2)+5x(x^3)+5x(2)+7(x^3)+7(2)}}}
{{{(f*g)(x)=f(x)*g(x)=2x^5+4x^2+5x^4+10x+7x^3+14}}}
{{{(f*g)(x)=f(x)*g(x)=2x^5+5x^4+7x^3+4x^2+10x+14}}}Rewrite terms in descending degree. Domain is all real numbers since the domain of the 2 equations is all real numbers. 




4.
This time you're going to divide the 2 equations
{{{(f/g)(x)=f(x)/g(x)=(2x^2+5x+7)/(x^3+2)}}}
{{{(f/g)(x)=f(x)/g(x)=(2x^2+5x+7)/(x^3+2)}}} And this is the simplest form since it cannot be reduced further. The domain is all real numbers except x cannot equal {{{-root(3,2)}}}