Find Polynomial Value Using Root Formation Method

 

❓ Question

Let $f(x)$ be a polynomial of degree 3 such that:

$$f(k)=-\frac{2}{k} \quad \text{for } k=2,3,4,5$$

Find the value of:

$$52-10f(10)$$

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🖼 Question Image

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✍️ Short Concept

This question uses:

👉 Polynomial construction
👉 Root formation trick
👉 Factor theorem.

Main idea:

Convert the given condition into a polynomial having known roots.

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🔷 Step 1 — Form a New Polynomial 💯

Given:

$$f(k)=-\frac{2}{k}$$

Multiply by $k$:

$$k f(k)+2=0$$

Define:

$$g(x)=x f(x)+2$$

Since $f(x)$ is degree 3,

$$g(x)$$

is degree 4.

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🔷 Step 2 — Identify Roots

For:

$$k=2,3,4,5$$

we get:

$$g(k)=0$$

So roots are:

$$2,3,4,5$$

Hence:

$$g(x)=a(x-2)(x-3)(x-4)(x-5)$$

That is:

$$x f(x)+2 = a(x-2)(x-3)(x-4)(x-5)$$

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🔷 Step 3 — Find Constant $a$

Put:

$$x=0$$
$$0\cdot f(0)+2 = a(-2)(-3)(-4)(-5)$$
$$2=120a$$
$$a=\frac1{60}$$

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🔷 Step 4 — Find $f(10)$

Put:

$$x=10$$
$$10f(10)+2 = \frac1{60}(8)(7)(6)(5)$$
$$10f(10)+2 = \frac{1680}{60}$$
$$10f(10)+2=28$$
$$10f(10)=26$$

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🔷 Step 5 — Final Calculation

$$52-10f(10) = 52-26$$
$$=26$$

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✅ Final Answer

$$\boxed{26}$$

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⭐ Golden JEE Insight

Whenever:

$$f(k)=\text{some value}$$

at multiple points,

try forming a new polynomial whose roots become those points.

This is one of the fastest polynomial tricks in JEE.