Question

What will be the H.C.F. of \[{x^3} - 1\] & \[{x^4} + {x^2} + 1\]
 A. \[x - 1\]
 B. \[{x^2} + x + 1\]
 C. \[{x^2} - x + 1\]
 D. \[{x^2} - x - 1\]

Answer 1

Hint: Factorise both the polynomials in the simplest form then compare to find the common factors and then find the product of the common factors to get the Highest Common Factor.

Complete step-by-step answer:
First let us try to factorise the polynomial \[{x^3} - 1\]
Clearly it is of the form \[{a^3} - {b^3}\] where \[a = x\] & \[b = 1\]

Now we know that
\[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\]
Substituting the values of a and b with x and 1 respectively. We get
\[ \Rightarrow {x^3} - 1 = (x - 1)({x^2} + x + 1)\]
So \[(x - 1)\] & \[({x^2} + x + 1)\] are the factors of the polynomial \[{x^3} - 1\]

For finding the factors of the next polynomial we will introduce some terms and then cancel out them by introducing the negative of it as well, from there we will try to take some commons and that will eventually led us to the factors we are looking for i.e.,
\[\begin{array}{l}
{x^4} + {x^2} + 1 = {x^4} + {x^3} + {x^2} - {x^3} - {x^2} - x + {x^2} + x + 1\\
 \Rightarrow {x^4} + {x^2} + 1 = {x^2}({x^2} + x + 1) - x({x^2} + x + 1) + ({x^2} + x + 1)\\
 \Rightarrow {x^4} + {x^2} + 1 = ({x^2} + x + 1)({x^2} - x + 1)
\end{array}\]

So \[({x^2} + x + 1)\] & \[({x^2} - x + 1)\] are the factors of the polynomial \[{x^4} + {x^2} + 1\]
If we try to compare the factors of the two polynomials we will get the Highest Common Factor as \[{x^2} + {x^{}} + 1\]

Therefore option B is the correct one.


Note: The factorisation of the second polynomial could have been done by a different approach i.e., we could have let \[y = {x^2}\] and then tried to find a quadratic equation in y and by sridharacharya method of solving quadratic equations, we can get the roots of the quadratic equation \[{y^2} + y + 1 = 0\] it is clear that the value of discriminant in this equation is less than 0 thus we will get two complex roots. Now by using the relation \[y = {x^2}\] the total number of roots will be four. Suppose the roots are denoted by \[\alpha ,\beta ,\chi ,\delta \] then the factors will be \[(x - \alpha ),(x - \beta ),(x - \chi ),(x - \delta )\] by finding the product of these factors we will again get the same factors as we got in the solution.