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 _________.

We have a cubic polynomial $$f(x)$$ that satisfies the four conditions

$$f(2)= -\dfrac{2}{2},\; f(3)= -\dfrac{2}{3},\; f(4)= -\dfrac{2}{4},\; f(5)= -\dfrac{2}{5}.$$

To make the right‐hand sides look simpler, notice that each value can be written as $$-\dfrac{2}{k}$$ when $$k=2,3,4,5.$$

Now define a new polynomial

$$g(x)=x\,f(x)+2.$$

This definition is chosen because, for any of the special points $$k=2,3,4,5,$$ we get

$$g(k)=k\,f(k)+2=k\left(-\dfrac{2}{k}\right)+2=-2+2=0.$$

Hence

$$g(2)=g(3)=g(4)=g(5)=0.$$

So, the numbers $$2,3,4,5$$ are roots of $$g(x).$$ Therefore $$g(x)$$ must be divisible by the quartic factor $$(x-2)(x-3)(x-4)(x-5).$$ Because $$g(x)$$ is itself of degree $$4$$ (it is $$x$$ times a cubic plus a constant), it can differ from that quartic only by a non-zero constant multiple. Thus,

$$g(x)=A\,(x-2)(x-3)(x-4)(x-5),$$

where $$A$$ is a constant yet to be determined.

To find $$A,$$ evaluate both sides at a convenient value of $$x$$ that is not a root. The simplest is $$x=0.$$ First compute $$g(0)$$ directly from its definition:

$$g(0)=0\cdot f(0)+2=2.$$

Next compute the right‐hand side at $$x=0$$:

$$(x-2)(x-3)(x-4)(x-5)\Big|_{x=0}=(-2)(-3)(-4)(-5).$$

Multiply step by step: $$(-2)(-3)=6,\; 6(-4)=-24,\; (-24)(-5)=120.$$ Hence

$$g(0)=A\cdot120.$$

Equating the two evaluations,

$$A\cdot120=2\quad\Longrightarrow\quad A=\dfrac{2}{120}=\dfrac{1}{60}.$$

Therefore, the explicit form of $$g(x)$$ is

$$g(x)=\dfrac{1}{60}(x-2)(x-3)(x-4)(x-5).$$

Remember that $$g(x)=x\,f(x)+2,$$ so we can solve for $$f(x):$$

$$f(x)=\dfrac{g(x)-2}{x}.$$

We are interested in $$f(10),$$ so substitute $$x=10:$$

$$f(10)=\dfrac{g(10)-2}{10}.$$

Compute $$g(10)$$ first:

$$g(10)=\dfrac{1}{60}(10-2)(10-3)(10-4)(10-5) =\dfrac{1}{60}\,(8)(7)(6)(5).$$

Multiply systematically: $$8\times7=56,\; 6\times5=30,\; 56\times30=1680.$$ Hence

$$g(10)=\dfrac{1680}{60}=28.$$

Now find $$f(10):$$

$$f(10)=\dfrac{28-2}{10}=\dfrac{26}{10}=\dfrac{13}{5}.$$

The problem asks for the value of $$52-10\,f(10).$$ Substitute $$f(10)=\dfrac{13}{5}:$$

$$52-10\,f(10)=52-10\left(\dfrac{13}{5}\right) =52-\dfrac{130}{5} =52-26 =26.$$

So, the answer is $$26$$.