anonymous one year ago Let f(x) = x^2 + x + 2 and g(x) = 2x^2 + 5. Find f(g(x)). Show each step of your work.

1. anonymous

@Vocaloid I just need to learn how to do the problem, not the answer.

2. Nnesha

replace x in f(x) function by g(x)

3. anonymous

I did that, I got: (2x^2+5)^2+2x^2+2

4. Nnesha

you forgot something $\huge\rm (2x^2+5)^2+\color{blue}{x}+2$ x = 2x^2 + 5

5. Nnesha

you forgot 5

6. anonymous

did I? Sorry.

7. anonymous

So to solve, is it 2x^4 or 4x^4

8. Nnesha

$\huge\rm (2x^2+5)^2+\color{blue}{2x^2 +5}+2$

9. Nnesha

$\huge\rm \color{reD}{x^2}+\color{blue}{x}+2$ $\huge\rm \color{ReD}{(2x^2+5)^2}+\color{blue}{2x^2 +5}+2$ x is 2x^2 + 5

10. anonymous

Yes, so now we have to square x.

11. Nnesha

yes( 2x^2 +5)^2 is same as (2x^2 +5) (2x^2 +5) so foil it

12. anonymous

$4x^4+20x^2+25$

13. anonymous

right?

14. Nnesha

yes that's right

15. anonymous

So now we combine like terms?

16. Nnesha

yesp

17. anonymous

$4x^4+20x^2+25+2x^2+5+2$

18. anonymous

turns into

19. anonymous

$4x^4+22x^2+32$

20. anonymous

Right?

21. Nnesha

yep

22. anonymous

So now I need to factor?

23. Nnesha

24. anonymous

Oh, is it? Good, my problem was I tried factoring that and had no idea how to do it.

25. anonymous

Thanks! :D

26. Nnesha

my pleasure and yes that's it they wants us to find f(g(x)) not factoring

27. anonymous

I fanned, so if I need anything, I'll make sure to tag you.

28. Nnesha

thanks i'll try my best :=)

29. anonymous

@Nnesha I have one more problem that seems pretty easy.

30. anonymous

I think I may know how to solve it but graphing and radicals are my weakpoint.

31. anonymous

Describe how to transform $(\sqrt[6]{x ^{5}})^{7}$ into an expression with a rational exponent

32. Nnesha

alright remember this exponent rule $\huge\rm \sqrt[m]{x^n} = x^\frac{ n }{ m }$ you can convert root to exponent form

33. anonymous

Yes I know this, it would be $(x^\frac{ 5 }{ 6 })^7$

34. anonymous

then you multiply 7 and 5?

35. Nnesha

36. Nnesha

$\huge\rm (x^m)^n = x^{ m \times n}$

37. anonymous

$x^\frac{ 35 }{ 6 }$

38. anonymous

thats a 35/6 btw

39. Nnesha

yep

40. anonymous

41. Nnesha

yep right

42. anonymous

K, thanks. Just wanted to check my answer.

43. Nnesha