Question

The GCD and LCM of two polynomials are \[x + 1\] and \[{x^6} - 1\] respectively. If one of the polynomials is \[{x^3} + 1\], find the other.

Answer 1

Hint: In the given question, we have been given the GCD (greatest common divisor, which is \[x + 1\]) and the LCM (least common multiple, which is \[{x^6} - 1\]) of two polynomials (out of which one is \[{x^3} + 1\]). We have to find the other unknown polynomial. This question can be solved by using the relationship between the GCD, LCM and the product of two numbers (or two polynomials). This says – the product of the GCD and LCM of two numbers (or two polynomials) is equal to their product. So, we just have to substitute in the values of the given quantities and find the unknown polynomial.

Formula Used:
We are going to use the formula of GCD, LCM and the product of two numbers, which is:
\[GCD \times LCM\] = product of the numbers

Complete step by step answer:
Given, GCD \[ = x + 1\]
LCM \[ = {x^6} - 1\]
First polynomial \[ = {x^3} + 1\]
Second polynomial \[ = y\] (to be calculated)
So, we just put in these values in the formula and evaluate the answer:
\[\left( {x + 1} \right)\left( {{x^6} - 1} \right) = \left( {{x^3} + 1} \right)\left( y \right)\]
We know, \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
So, we have:
\[\dfrac{{\left( {x + 1} \right)\left( {{x^3} - 1} \right)\left( {{x^3} + 1} \right)}}{{\left( {{x^3} + 1} \right)}} = y\]
Solving the expression, we have:
\[y = \left( {x + 1} \right)\left( {{x^3} - 1} \right)\]
or, \[y = {x^4} + {x^3} - x - 1\]

Hence, the other polynomial is \[{x^4} + {x^3} - x - 1\].

Additional Information:
In this question, we solved the expression by substituting the value of \[{x^6} - 1\]. Though it is more lengthy and quite complicated, we could have solved the expression by dividing the \[{x^3} + 1\] in the denominator by the \[x + 1\] in the numerator. It takes more time as we then would need to divide the \[{x^6} - 1\] term by the quotient of \[{x^3} + 1\] and \[x + 1\], which is \[\left( {{x^2} + 1 - x} \right)\]. And that is why we find the most effective way of solving the expression.

Note:
Here, we saw that we just needed to make one simple substitution of the formula of the variable’s sixth power. We cannot just multiply the two terms in the numerator, then divide by denominator; this all makes the work unnecessarily difficult and is very time-consuming. We must know all the basic formulae, like here we saw that the difference of two squares’ formula was involved in solving this question. Simply just multiplying the two polynomials and then dividing it with the given polynomial is no good, as it is very lengthy, messy, and most of all, very time consuming. So, we must find the hacks to solve our problem effectively.