Question

What is the HCF of the polynomials \[{x^4} - 3x + 2,{x^3} + - 3{x^2} + 3x - 1{\text{ and }}{x^4} - 1\]?
A. \[x - 1\]
B. \[x + 2\]
C. \[{x^2} - 1\]
D. None of these

Answer 1

Hint: First of all, write the factors of the given polynomials separately by using the algebraic identities. The highest common factor is found by multiplying all the factors which appear in the given polynomials. So, use this concept to reach the solution of the given problem.

Complete Step-by-Step solution:
The given polynomials are \[{x^4} - 3x + 2,{x^3} + - 3{x^2} + 3x - 1{\text{ and }}{x^4} - 1\]
Now, consider the factors of \[{x^4} - 3x + 2\]
Adding and subtracting \[{x^3} + {x^2}\] on both sides
\[
  {x^4} - 3x + 2 = {x^4} + {x^3} + {x^2} - {x^3} - {x^2} + 3x + 2 \\
  {x^4} - 3x + 2 = {x^4} - {x^3} + {x^3} - {x^2} + {x^2} - x - 2x + 2 \\
\]
Taking the terms in common, we have
\[
  {x^4} - 3x + 2 = {x^3}\left( {x - 1} \right) + {x^2}\left( {x - 1} \right) + x\left( {x - 1} \right) - 2\left( {x - 1} \right) \\
  \therefore {x^4} - 3x + 2 = \left( {x - 1} \right)\left( {{x^3} + {x^2} + x - 2} \right) \\
\]
Now, consider the factors of \[{x^3} - 3{x^2} + 3x - 1\]
\[
  {x^3} - 3{x^2} + 3x - 1 = {\left( {x - 1} \right)^3} \\
  \therefore {x^3} - 3{x^2} + 3x - 1 = \left( {x - 1} \right){\left( {x - 1} \right)^2} \\
\]
And, the factors of \[{x^4} - 1\]
\[{x^4} - 1 = {x^4} - {1^4}\]
We know that \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\]
\[
  {x^4} - 1 = \left( {{x^2} - {1^2}} \right)\left( {{x^2} + {1^2}} \right) \\
  \therefore {x^4} - 1 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right) \\
\]
As HCF is the highest common factor, the HCF of given polynomials is \[x - 1\].
Thus, the correct option is A. \[x - 1\]

Note: Here the algebraic identities used are \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\]. As we have obtained only one common factor in all the given polynomials, this common factor is our required HCF.