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# help urgent

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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.)

Jan 17, 2020

#1
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pls im confused

Jan 17, 2020
#2
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I worked through this

f(x)  =  (-3/2)x^2  + (1/2)x + 4

g (x) = (f(x))^16

Here is the expansion of this

(43046721 x^32)/65536 - (14348907 x^31)/4096 - (157837977 x^30)/8192 + (518154975 x^29)/4096 + (4068180855 x^28)/16384 - (8827766451 x^27)/4096 - (14523632991 x^26)/8192 + (94257891351 x^25)/4096 + (220712269635 x^24)/32768 - (706381131105 x^23)/4096 - (23495103567 x^22)/8192 + (3940745861637 x^21)/4096 - (1959573354921 x^20)/16384 - (16925275164105 x^19)/4096 + (6150417128055 x^18)/8192 + (57033142395909 x^17)/4096 - (166621058538623 x^16)/65536 - (19011047465303 x^15)/512 + (683379680895 x^14)/128 + (626862043115 x^13)/8 - (24192263641 x^12)/4 - 129736489272 x^11 - 1031335136 x^10 + 165371348480 x^9 + 17223697920 x^8 - 156919300096 x^7 - 32238338048 x^6 + 104507375616 x^5 + 32107397120 x^4 - 43620761600 x^3 - 17716740096 x^2 + 8589934592 x + 4294967296

Adding the  coefficients on the even powered terms  would be very tedious....as you can imagine  !!!

Jan 17, 2020
#3
+109739
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I WILL be glad to show you how I got f(x)....but....adding those co-efficients.....forget it  !!!!

Jan 17, 2020
#4
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can u tho?

i like to know things

Guest Jan 17, 2020
#5
+109739
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We have  this system of equations

a(-4)^2  + b(-4)  + c  = -22

a(-1)^2  + b(-1) + c = 2

a(2)^2 + b(2) + c  =  -1      simplify these

16a -4b + c  = -22

a  - b   +  c =  2

4a  + 2b  + c  = -1

Subtract the  second and third equations  from the  first and we get that

15a - 3b  = -24     ⇒   5a -b  = -8     ⇒   -10a + 2b = 16

12a  -6b  = -21   ⇒    4a - 2b  = -7   ⇒      4a  - 2b  = -7

Add the last two equations  and we have that

-6a = 9      divide both sides by -6

a  =9/-6 = -3/2

And

5(-3/2) - b  = -8

-15/2 - b = -16/2

-b = -16/2 + 15/2

-b = -1/2

b = 1/2

And

a - b + c  = 2

(-3/2) - (1/2) + c  = 2

-2 + c  = 2

c   = 4

So

f (x)   = ax^2 + bx + c  =  (-3/2)x^2 + (1/2)x  + 4

Jan 17, 2020