Let f(x) be a polynomial of degree 3 such tha | vedclass

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(A) Given $f(k) = -\frac{2}{k}$,we can write $k f(k) + 2 = 0$ for $k = 2, 3, 4, 5$.Let $g(x) = x f(x) + 2$. Since $f(x)$ is a polynomial of degree $3$

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$26$

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(A) Given $f(k) = -\frac{2}{k}$,we can write $k f(k) + 2 = 0$ for $k = 2, 3, 4, 5$.Let $g(x) = x f(x) + 2$. Since $f(x)$ is a polynomial of degree $3$,$g(x)$ is a polynomial of degree $4$.Since $g(k) = 0$ for $k = 2, 3, 4, 5$,we can write $g(x) = \lambda(x-2)(x-3)(x-4)(x-5)$ for some constant $\lambda$.To find $\lambda$,consider the coefficient of $x^3$ in $f(x)$. Since $g(x) = x f(x) + 2$,the coefficient of $x^4$ in $g(x)$ is the same as the coefficient of $x^3$ in $f(x)$.From $g(x) = \lambda(x^4 - 14x^3 + \dots)$,we have $\lambda = 	ext{leading coefficient of } f(x)$.However,we can find $\lambda$ by evaluating at $x=0$: $g(0) = 0 \cdot f(0) + 2 = 2$.So,$2 = \lambda(0-2)(0-3)(0-4)(0-5) = \lambda(120)$,which gives $\lambda = \frac{2}{120} = \frac{1}{60}$.Now,$10 f(10) + 2 = g(10) = \frac{1}{60}(10-2)(10-3)(10-4)(10-5) = \frac{1}{60}(8)(7)(6)(5) = \frac{1680}{60} = 28$.Thus,$10 f(10) = 28 - 2 = 26$.Finally,$52 - 10 f(10) = 52 - 26 = 26$.

Let $f(x)$ be a polynomial of degree $3$ such that $f(k) = -\frac{2}{k}$ for $k = 2, 3, 4, 5$. Then the value of $52 - 10 f(10)$ is equal to:

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