Find the composition of f(g(x)), when
f(x)=7x^2-2x+9 and
g(x)=x-4
Simplify

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- Jamierox4ev3r

Find the composition of f(g(x)), when
f(x)=7x^2-2x+9 and
g(x)=x-4
Simplify

- schrodinger

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- Jamierox4ev3r

I have a pretty clear idea of how to go about doing this, so I'll just show all of my process. I'll also point out the one small detail that I had a bit of doubt on.

- Jamierox4ev3r

\(7(x-4)^{2}-2(x-4)+9\)
\(7(x-4)(x-4)-2x+8+9\)
\((7x-28)(x-4)-2x+17\)
\(7x^{2}-28x-28x+112-2x+17\)
\(7x^{2}-58x+129\)
^^
I believe this is correct...but I forget the rules on difference of squares.
does (x-2)^2 turn into (x-2)(x-2) or (x-2)(x+2) ??
Also, do I need to factor the final answer?

- madhu.mukherjee.946

try and see if it can be done,then its good

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## More answers

- Jamierox4ev3r

see it what can be done?

- madhu.mukherjee.946

i don't think it can be factored

- jim_thompson5910

x^2 means x times x
anything squared is that number multiplied by itself
so (x-2)^2 means (x-2) times (x-2)

- jim_thompson5910

\[\Large 7x^{2}-58x+129\] is the correct final answer

- Jamierox4ev3r

Oh. You're referring to my question on whether it could be factor or not.
oh and @jim_thompson5910 , so in the problem, if you look at my work, my assumption was indeed that (x-2)^2 was (x-2)(x-2). Thank you! I thought I remembered some funky rule from early high school about difference of squares or something like that. So I was correct all along. Wow I overthought this. Thank again!

- jim_thompson5910

you're thinking of a^2 - b^2 = (a-b)(a+b)

- madhu.mukherjee.946

its perfectly all right

- madhu.mukherjee.946

u can solve it by the formula

##### 1 Attachment

- Jamierox4ev3r

Oh okay. These rules are so easy to get confused, especially when trying to recall from memory

- madhu.mukherjee.946

this formula is known as the SRIDHAR ACHARYA formula

- madhu.mukherjee.946

don't worry by this formula it will get easy

- madhu.mukherjee.946

put 7x^-58x+129 in this formula

- Jamierox4ev3r

Don't you mean the quadratic formula? And yes, I know that you can use the quadratic formula to factor. Problem is, I see now that when I use the quadratic formula, the roots are imaginary. Therefore, This cannot be factored any further. Jim's answer is the final answer

- madhu.mukherjee.946

yeah

- Jamierox4ev3r

@madhu.mukherjee.946 thanks for trying to help, but I fully understand now. I've used the quadratic formula before, and I understand what it does \(\ddot\smile\)

- madhu.mukherjee.946

okay.your welcome

- jim_thompson5910

You can use the discriminant formula as a shorter way to see if you can factor or not
if D = b^2 - 4ac is a perfect square and a positive number, then you can factor

- Jamierox4ev3r

Woah, that's a helpful tip. *cough* I have a program in my calculator that solves all quadratics for me, so that's how I knew that the roots would be imaginary. But when in a pinch, when without a calculator, this is an extremely convenient tip. Thanks for bringing that back into my memory

- jim_thompson5910

definitely a handy thing to have

- jim_thompson5910

you're welcome

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