UPSKILL MATH PLUS
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Learn moreConsider 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 results in "the product of any two polynomials are 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.
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