jfan17

I'm responding to your posts because we are not here to fix your lousy formatting, we're here to help you with the problems. If you want me to continue helping you out, I'd suggest you to not copy paste directly from whatever site you're using and cheating off of.

Name \(f(x) = ax^2 + bx + c\)

\(f(-4) = -22 \)

\(f(-1) = 2\)

\(f(2) = -1\)

Substituting -4 for x in our first equation, we get:

\(16a -4b + c = -22\)

Substituting -1 for x in our second equation, we get:

\(a-b+c = 2\)

Finally, substituting 2 for our third equation, we get:

\(4a + 2b + c = -1\)

First we can subtract equations (3) and (2) to get:

\(3a +3b = -3\)

Dividing by 3 on both sides, we then get:

\(a+b = -1\)

\(a = -1 -b\)

Substitute into our first equation, we get:

\(-16-16b-4b+c = -22\)

\(-16-20b + c = -22\)

Substitute back into our second equation, we get:

\(-1-2b + c = 2\)

Subtracting our first equation from our second equation, we then arrive at:

\(15+18b =24\)

\(18b = 9 \)

\(b = \frac1{2}\)

Now that we have b, we're looking for a and c.

If we substitute back into our equation

\(a+b = -1\)

\(a + \frac1{2} = -1\)

\(a = -\frac3{2}\)

Substitute into any of our equations(I'll use the second equation for simplicity)

\(-\frac3{2} -\frac1{2} + c = 2\)

\(-2 + c = 2 \)

\(c = 4\)

We have the quadratic:

\(f(x) = -\frac3{2}x^2 + \frac1{2}x + 4\)

We need to take the 16th power of this quadratic. 

g(x) = f(x)¹⁶

We can write :

g(x) = a₀ + a₁x + a₂x² +..........

If we plug in 1 into this equation, we get:

g(1) = a₀ + a₁ + a₂ +..........

3¹⁶ = a₀ + a₁ + a₂ +..........

If we plug in -1 into the function g(x), we get:

g(-1) = a₀ - a₁ + a₂ +..........

2¹⁶ = a₀ - a₁ + a₂ +..........

If we add these two equations, we see that all of the odd coefficients cancel out, leaving us with only the even ones. We then get,

43112257 = 2(a₀ + a₂ +..........)

Our answer is then 43112257/2.

No problem!

Also please make sure to properly format your question! I'd check this site out for reference. It'll help you understand some properties and conditions of fractional modulo

https://math.stackexchange.com/questions/864568/is-it-possible-to-do-modulo-of-a-fraction

Sorry I was going to post an answer to this, but it got deleted when I tried submitting :(. I only got through part of this, but I'll try uploading a new answer tomorrow. In the meantime, try searching up "bezouts identity" and its relation with fractional modulo. What can you immediately realize won't satisfy the requirements of the given expression?(Hint: it has something to do with the denominators being coprime to the modulus). 

:/

https://www.quora.com/How-many-zeroes-do-we-write-when-we-write-all-the-integers-from-1-to-243-in-base-3

@EpicWater

This is a reference website, has the same (albeit more detailed) solution that CGT outlined for us

No it would be the square root of 1762,

\(\sqrt{9^2 + 41^2} = \sqrt{81 +1681} = \sqrt{1762}\)

format! format! format please! for all the latex parts, just paste it into the latex editor. There's one built into the answer box for web2.0calc. We're not here to translate LaTeX into a problem!

A simple and concise method is to add all 3 of these equations up. We don't have to subtract any of them to find individual variables. If we add all 3 of these, we get:

2a + 2b + 2c = 56

If we divide by 2(factor out 2) from both sides, we then get:

a + b + c = 28

See? Not so bad, was it?

Side by side comparison?

https://web2.0calc.com/api/ssl-img-proxy?src=%2Fimg%2Fupload%2Fbc435091ccbca0f0af8a1387%2Fcphillanswer.png&l=1

you changed one thing, and that's the brackets in this part alone. I could compare other parts, but this is just a petty argument in general. If you know you copied someone else's work, that's ok, just make sure to credit them. That's my point here. End of discussion.