let f(x)=2x+2, g(x)=x^2 and h(x) 3/x-2
What is f(x)+g(x), g(x)/f(x), h(x)-f(x), and h(x) * g(x).

Question

let f(x)=2x+2, g(x)=x^2 and h(x) 3/x-2
What is f(x)+g(x), g(x)/f(x), h(x)-f(x), and h(x) * g(x).

zepdrix · Answer

$$\large\rm \color{royalblue}{f(x)=2x+2},\qquad \color{green}{g(x)=x^2},\qquad \color{orangered}{h(x)=\frac{3}{x-2}}$$So our first problem to figure out is$$\large\rm \color{royalblue}{f(x)}+\color{green}{g(x)}$$

Any ideas?

anonymous · Answer

The base numer is different. If the x^2 is x will be easier. Because of that, I tried to make them have the same base number, but I feel little bit lost today. I need so ideas about this.

zepdrix · Answer

The ..."base number" is different, yes.
Which means they cannot be combined together.
$$\large\rm \color{royalblue}{f(x)}+\color{green}{g(x)}$$Which means the first one is as simple as this,$$\large\rm \color{royalblue}{2x+2}+\color{green}{x^2}$$And that would be your final answer for the first one.

zepdrix · Answer

$$\large\rm \frac{\color{green}{g(x)}}{\color{royalblue}{f(x)}}\quad=\frac{\color{green}{x^2}}{\color{royalblue}{2x+2}}$$Just plug in the matching colors.
Again, no need to simplify.
You `could` do polynomial long division but that will leave you with a remainder. This is a good answer right here.

anonymous · Answer

That's enough. I can do the rest of them. Thanks so much. :)

anonymous · Answer

Before, I just not sure what should be the good answer. Now, I got it. Thanks again.

zepdrix · Answer

cool :)