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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