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Let \(p(x)\) be a monic, quartic polynomial, such that \(p(1) = 3, p(3) = 11, \)  and \(p(5) = 27.\) Find \(p(-2) + 7p(6).\)
 

 

I don't know how to find p(6) or p(-2), or if I even have to find them to solve the question.

 

Can someone help me? Thank you!

 Dec 14, 2020
 #1
avatar+419 
+1

Monic indicates there is only one variable used, so we can say that $p(x) = ax^4 + bx^3 + cx^2 + dx + f,$ where $a,b,c,d,$ and $f$ are the coefficients. It appears to be that when we input $x$ as $1$, we just get $a + b + c + d + f = 3,$ and with $3$, we get $81a + 27b + 9c + 3d + f = 11,$ and with $5$ we have $625a + 125b + 25c + 5d + f = 27. $

 

We are trying to solve for $16a - 8b + 4c - 2d + f + 7 \cdot (1296a + 216b + 36c + 6d + f).$

 

See if you can simplify and/or find out this equation. Another thing you might notice is that if we have p(x) = y, then $\overline{abcdf}$ is a 5 digit number in base x, which equals y in base 10!

 Dec 14, 2020
 #2
avatar
+1

x^2 + 2

 

so  p(-2)  + 7 p(6)

 =     6     + 266= 272

 Dec 14, 2020
 #3
avatar+419 
+1

it says quartic, that's why I was skeptical about that solution.

Pangolin14  Dec 14, 2020

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