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# I'm really stumped... (polynomials due very soon)

-1
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Let f(x) be a quadratic polynomial such that $$f(-4) = -22,f(-1)=2 \text { and } f(2)=-1.$$ Let $$g(x) = f(x)^{16}.$$ Find the sum of the coefficients of the terms in g(x) that have even degree.

P.S. I'd like to know how to get the equation for f(x), I'm still stuck here... (it would be really helpful to know the steps).

Mar 29, 2020
edited by Guest  Mar 29, 2020

#1
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You don't need to find the explicit form of f(x)16 here.  See the following: Mar 29, 2020
edited by Alan  Mar 29, 2020
edited by Alan  Mar 29, 2020
edited by Alan  Mar 29, 2020
#2
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Thanks so much! do you mind telling me how you got the values for a, b and c? I was really stumped there.

Guest Mar 29, 2020
#3
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Write the three equations as:

-22 = 16a - 4b + c.  (1)

2 = a - b + c.           (2)

-1 = 4a + 2b + c.     (3)

(3) - (2):  3a + 3b = -3.  (4)

(1) - (2):  15a - 3b = -24  (5)

(4) + (5):  18a = -27.   Divide by 9:     2a = -3.     a = -3/2

Substitute for a in (4) or (5) to find b, then substitute both into any of (1), (2) or (3) to find c.

Alan  Mar 29, 2020
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Just one more thing, why did you do $$\frac{f(1)^{16}+f(-1)^{16}}2$$?

Guest Mar 30, 2020
#5
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Let g(x) = f(x)16.     Write g(x) as  g(x) = g(x)eventerms + g(x)oddterms.

g(1) = g(1)eventerms + g(1)oddterms  ...(1)

g(-1) = g(-1)eventerms + g(-1)oddterms

g(-1) =  g(1)eventerms - g(1)oddterms  ...(2)

(1) + (2):  g(1) + g(-1) = 2*g(1)eventerms

so: g(1)eventerms = (g(1) + g(-1))/2

or g(1)eventerms = (f(1)16 + f(-1)16)/2

Alan  Mar 30, 2020