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# Functions

0
55
3
+156

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
+428
+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
+1

x^2 + 2

so  p(-2)  + 7 p(6)

=     6     + 266= 272

Dec 14, 2020
#3
+428
+1

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

Pangolin14  Dec 14, 2020