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Let $f$ be a cubic polynomial such that $f(0) = 0$, $f(1) = 0$, $f(2) = 0$, and $f(3)=0$. What is the sum of the coefficients of $f$?

 Nov 30, 2023
 #1
avatar+4 
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Let's consider the cubic polynomial $f(x)$ with the given conditions: $f(0) = 0$, $f(1) = 0$, $f(2) = 0$, and $f(3) = 0$.

Since $f(0) = 0$, we can conclude that $x = 0$ is a root of the polynomial. Similarly, $f(1) = 0$ implies that $x = 1$ is a root, $f(2) = 0$ implies that $x = 2$ is a root, and $f(3) = 0$ implies that $x = 3$ is a root.

Using this information, we can express the polynomial $f(x)$ in factored form as:

\[f(x) = k(x - 0)(x - 1)(x - 2)(x - 3)\]

where $k$ is some constant.

To find the sum of the coefficients of $f(x)$, we can expand the polynomial and look at the coefficient of $x^0$ (the constant term), $x^1$, $x^2$, and $x^3$.

Expanding the polynomial:

\[f(x) = k(x^4 - 6x^3 + 11x^2 - 6x)\]

The sum of the coefficients of $f(x)$ is given by:

\[\text{Sum of coefficients} = k(1 - 6 + 11 - 6) = k\]

Since we are only interested in the sum of the coefficients, we don't need to determine the exact value of $k$. From the given conditions, we know that $f(x)$ has four roots, which means it is a cubic polynomial. Hence, the sum of the coefficients of $f(x)$ is equal to $k$.

Therefore, the sum of the coefficients of $f(x)$ is $k$.

 Dec 1, 2023
 #2
avatar+33595 
0

A cubic polynomial has three roots.  A quartic polynomial has four.

 Dec 1, 2023

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