- anonymous

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

- jamiebookeater

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- anonymous

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

- Nnesha

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

- anonymous

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

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## More answers

- Nnesha

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

- Nnesha

you forgot 5

- anonymous

did I? Sorry.

- anonymous

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

- Nnesha

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

- 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

- anonymous

Yes, so now we have to square x.

- Nnesha

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

- anonymous

\[4x^4+20x^2+25\]

- anonymous

right?

- Nnesha

yes that's right

- anonymous

So now we combine like terms?

- Nnesha

yesp

- anonymous

\[4x^4+20x^2+25+2x^2+5+2\]

- anonymous

turns into

- anonymous

\[4x^4+22x^2+32\]

- anonymous

Right?

- Nnesha

yep

- anonymous

So now I need to factor?

- Nnesha

nope that's ur answer

- anonymous

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

- anonymous

Thanks! :D

- Nnesha

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

- anonymous

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

- Nnesha

thanks i'll try my best :=)

- anonymous

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

- anonymous

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

- anonymous

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

- Nnesha

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

- anonymous

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

- anonymous

then you multiply 7 and 5?

- Nnesha

yep! right you already know

- Nnesha

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

- anonymous

\[x^\frac{ 35 }{ 6 }\]

- anonymous

thats a 35/6 btw

- Nnesha

yep

- anonymous

And that is my answer?

- Nnesha

yep right

- anonymous

K, thanks. Just wanted to check my answer.

- Nnesha

your answer is correct you're a mathematician :=)

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