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### Theory:

Consider two numbers, $$15$$ and $$20$$.

The LCM of $$15$$ and $$20$$ is $$60$$. That is, $$LCM(15,20) = 60$$.

The GCD of $$15$$ and $$20$$ is $$5$$. That is, $$GCD(15,20) = 5$$.

Now, $$LCM(15,20) \times GCD(15,20) = 60 \times 5 = 300$$

$$\Rightarrow LCM(15,20) \times GCD(15,20) = 15 \times 20$$

This gives us the result that "the product of any two polynomials is equal to the product of their LCM and GCD".

That is, $$f(x) \times g(x) = LCM[f(x),g(x)] \times GCD[f(x),g(x)]$$.

Let us understand the concept using an example.
Example:
Let $$f(x) = 21(x^4 - x^2)$$ and $$g(x) = 16(x^2 + 3x)^2$$. Let us verify $$f(x) \times g(x) = LCM[f(x),g(x)] \times GCD[f(x),g(x)]$$.

Solution:

To prove: $$f(x) \times g(x) = LCM[f(x),g(x)] \times GCD[f(x),g(x)]$$.

Proof: $$f(x) = 21(x^4 - x^2) = 3 \times 7 \times x^2 \times (x^2 - 1) = 3 \times 7 \times x^2 \times (x + 1)(x - 1)$$

$$g(x) = 16(x^2 + 3x)^2 = 2^4 \times (x^4 + 6x^3 + 9x^2) = 2^4 \times x^2 \times (x^2 + 6x + 9) = 2^4 \times x^2 \times (x + 3)(x + 3)$$

Now, $$LCM[f(x),g(x)] = 3 \times 7 \times 2^4 \times x^2 \times (x + 1)(x - 1) \times (x + 3)(x + 3)$$

$$= 336 \times x^2(x^2 -1)(x + 3)^2$$

$$GCD[f(x),g(x)] = x^2$$

Consider the $$LHS = f(x) \times g(x)$$

$$f(x) \times g(x) = 21(x^4 - x^2) \times 16(x^2 + 3x)^2 = 336(x^4 - x^2)(x^2 + 3x)^2$$ ---- ($$1$$)

Consider the $$RHS = LCM[f(x),g(x)] \times GCD[f(x),g(x)]$$

$$LCM[f(x),g(x)] \times GCD[f(x),g(x)] = 336 \times x^2(x^2 - 1)(x + 3)^2 \times x^2$$

$$= 336x^2(x^2 -1) \times x^2(x^2 + 6x + 9)$$

$$= 336(x^4 - x^2)(x^4 + 6x^3 + 9x^2)$$

$$= 336(x^4 - x^2)(x^2 + 3x)^2$$ ---- ($$2$$)

From equations ($$1$$) and ($$2$$), we have:

$$f(x) \times g(x) = LCM[f(x),g(x)] \times GCD[f(x),g(x)]$$

Hence, we proved.