The GCD and LCM of two polynomials are x+1 and x^6-1 respectively. if one of the polynomials is x^3+1find others
Answers
Given :-
The GCD and LCM of two polynomials are x+1 and x⁶-1.
One of the polynomials is x³+1.
To find :-
The other polynomial.
Solution :-
Given that
One of the polynomials is x³+1
Let the other polynomial be a
The GCD of the polynomials = x+1
The LCM of the polynomials = x⁶-1
We know that
The product of the HCF and LCM of two numbers is equal to the product of the two numbers
=> LCM × GCD = Product of two polynomials
=> (x⁶-1)(x+1) = (x³+1)×a
It can be written as
=> [(x³)²-1²](x+1) = (x³+1)×a
=> (x³+1)(x³-1)(x+1) = (x³+1)×a
Since, (a+b)(a-b) = a²-b²
=> [(x³+1)(x³-1)(x+1)]/(x³+1) = a
=> a = [(x³+1)(x³-1)(x+1)]/(x³+1)
=> a = (x³-1)(x+1)
=> a = x³(x+1)-1(x+1)
=> a = x⁴+x³-x-1
Answer :-
The other polynomial is x⁴+x³-x-1
Used formulae:-
• The product of the HCF and LCM of two numbers is equal to the product of the two numbers
• (a+b)(a-b) = a²-b²
Step-by-step explanation:
Given :-
The GCD and LCM of two polynomials are x+1 and x⁶-1.
One of the polynomials is x³+1.
To find :-
The other polynomial.
Solution :-
Given that
One of the polynomials is x³+1
Let the other polynomial be a
The GCD of the polynomials = x+1
The LCM of the polynomials = x⁶-1
We know that
The product of the HCF and LCM of two numbers is equal to the product of the two numbers
=> LCM × GCD = Product of two polynomials
=> (x⁶-1)(x+1) = (x³+1)×a
It can be written as
=> [(x³)²-1²](x+1) = (x³+1)×a
=> (x³+1)(x³-1)(x+1) = (x³+1)×a
Since, (a+b)(a-b) = a²-b²
=> [(x³+1)(x³-1)(x+1)]/(x³+1) = a
=> a = [(x³+1)(x³-1)(x+1)]/(x³+1)
=> a = (x³-1)(x+1)
=> a = x³(x+1)-1(x+1)
=> a = x⁴+x³-x-1
Answer :-
The other polynomial is x⁴+x³-x-1
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