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Suppose f is a polynomial such that f(0)=47, f(1)=32,f(2)=-13, and f(3)=16. what is the sum of the coefficients of f?Asked by Lily
Suppose f is a polynomial such that f(0)=47, f(1)=32, f(2)=-13, and f(3)=16. What is the sum of the coefficients of f?
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Answered by
Steve
assuming that f(x) = ax^3+bx^2+cx+d,
d=47
a+b+c+47 = 32
8a+4b+2c+47 = -13
27a+9b+3c+47 = 16
f(x) = (52x^3 - 201x^2 + 104x + 47)/3
d=47
a+b+c+47 = 32
8a+4b+2c+47 = -13
27a+9b+3c+47 = 16
f(x) = (52x^3 - 201x^2 + 104x + 47)/3
Answered by
Anonymous
The sum of the coefficients of any polynomial $f(x)$ equals $f(1)$, since letting $x = 1$ in any term $cx^n$ gives us the coefficient $c$. That is, if
\[f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0,\]
then
\begin{align*}
f(1) &= a_n\cdot 1^n + a_{n-1}\cdot1^{n-1} + \cdots + a_1\cdot 1 + a_0\\
&=a_n+a_{n-1} + \cdots + a_1 + a_0.
\end{align*}
So, the sum of the coefficients of $f$ is $f(1) = \boxed{32}$.
\[f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0,\]
then
\begin{align*}
f(1) &= a_n\cdot 1^n + a_{n-1}\cdot1^{n-1} + \cdots + a_1\cdot 1 + a_0\\
&=a_n+a_{n-1} + \cdots + a_1 + a_0.
\end{align*}
So, the sum of the coefficients of $f$ is $f(1) = \boxed{32}$.
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