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Let \(f(x)\) be a quadratic polynomial such that \(f(-4) = -22, f(-1)=2\), and \(f(2)=-1\). Let \(g(x) = f(x)^{16}\). Find the sum of the coefficients of the terms in \(g(x)\) that have even degree. (For example, the sum of the coefficients of the terms in \(-7x^3 + 4x^2 + 10x - 5\) that have even degree is \((4) + (-5) = -1.\))

 Mar 24, 2020
 #1
avatar+485 
+2

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 + a2x+..........

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.

 Mar 24, 2020
edited by jfan17  Mar 24, 2020
edited by jfan17  Mar 24, 2020
edited by jfan17  Mar 24, 2020
edited by jfan17  Mar 24, 2020
 #2
avatar+4 
+1

How did you figure out that the polynomial had a degree of 2?

ffffffffffffffffff  Mar 24, 2020
 #3
avatar+485 
+1

It says it's a quadratic, which means it has a degree of 2 if that answers your question. "Let f(x) be a quadratic polynomial"

jfan17  Mar 24, 2020
edited by jfan17  Mar 24, 2020
 #4
avatar+4 
+1

Sorry. I'm a bit brain-dead today.

ffffffffffffffffff  Mar 24, 2020
 #5
avatar+485 
+1

No worries!  :)

jfan17  Mar 24, 2020
 #6
avatar+20 
+1

I understand everything except when you raised it all to the sixteenth power, do you mind elaborating further on how you got to 

\(g(x) = a_0 + a_1x + a_2x^2 +..........\)?

 

smiley

Joshy1028  Mar 26, 2020
 #7
avatar+485 
+1

I just wrote out g(x)'s polynomial form. In other words, because I didn't know the actual coefficients, I substituted with variables like a0, a1, etc. to denote the coefficient for that x term

jfan17  Mar 26, 2020

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