Guest Dec 4, 2014

#1**+10 **

Suppose is a polynomial such that , , , and . What is the sum of the coefficients of ?

This polynomial has - at least - two real zeroes.....(there may be more).....let's suppose that it's a cubic

So we have

f(x) = ax^3 + bx^2 + cx + d ....Note, "d" has to be equal to 47

So we have this system

a(1)^3 +b(1)^2 + c(1) + 47 = 32

a(2)^3 +b(2)^2 + c(2) + 47 = -13

a(3)^3 +b(3)^2 + c(2) + 47 = 16

We can simplify these to:

a + b + c = -15

8a + 4b + 2c = -60

27a + 9b + 3c = -31

I used WolframAlpha solve this one .... (I'm lazy.....)

a = 52/3 b = -67 c = 104/3 d = 47..... I'll let you sum these.......!!!!

Here's the graph.........https://www.desmos.com/calculator/gnbrhqhxdr

(Yep...that works !!!!)

Note that - if this is a cubic - there * had* to be another real zero...also, this solution may not be unique....other functions might be possible......

CPhill
Dec 4, 2014

#1**+10 **

Best Answer

Suppose is a polynomial such that , , , and . What is the sum of the coefficients of ?

This polynomial has - at least - two real zeroes.....(there may be more).....let's suppose that it's a cubic

So we have

f(x) = ax^3 + bx^2 + cx + d ....Note, "d" has to be equal to 47

So we have this system

a(1)^3 +b(1)^2 + c(1) + 47 = 32

a(2)^3 +b(2)^2 + c(2) + 47 = -13

a(3)^3 +b(3)^2 + c(2) + 47 = 16

We can simplify these to:

a + b + c = -15

8a + 4b + 2c = -60

27a + 9b + 3c = -31

I used WolframAlpha solve this one .... (I'm lazy.....)

a = 52/3 b = -67 c = 104/3 d = 47..... I'll let you sum these.......!!!!

Here's the graph.........https://www.desmos.com/calculator/gnbrhqhxdr

(Yep...that works !!!!)

Note that - if this is a cubic - there * had* to be another real zero...also, this solution may not be unique....other functions might be possible......

CPhill
Dec 4, 2014