What is an irreducible polynomial?
Answer
521.1k+ views
Hint: To know what is an irreducible polynomial first we should know what is a polynomial. Then using the definition of types of polynomials we can get the required answer.
Complete step by step answer:
A polynomial with integer coefficients that cannot be factored into polynomials of lower degree, also with integer coefficients or the polynomial which is not factored at all is called an irreducible polynomial.
Examples: ${x^2} + x + 1$
The given polynomial is an irreducible polynomial. There is no way to find two integers a and b such that their product is $1$ and their sum is also $1$, so we cannot factor into linear terms $(x + a)\,(x + b)$.
${x^2} + 1$
The given polynomial is an irreducible polynomial. It has no simpler factors with integer coefficients.
${x^2} - 2$
The given polynomial is an irreducible polynomial. It has no simpler factors with integer coefficients. However, we could factor it as $(x - \sqrt 2 )\,(x + \sqrt 2 )$ but we are allowed to use irrational numbers.
Note:
Irreducibility of a polynomial depends on the coefficient numbers. To show that whether the given polynomial is irreducible or not we can use the long division method. If a polynomial with degree $2$ or higher is irreducible, then it has no real roots.
Complete step by step answer:
A polynomial with integer coefficients that cannot be factored into polynomials of lower degree, also with integer coefficients or the polynomial which is not factored at all is called an irreducible polynomial.
Examples: ${x^2} + x + 1$
The given polynomial is an irreducible polynomial. There is no way to find two integers a and b such that their product is $1$ and their sum is also $1$, so we cannot factor into linear terms $(x + a)\,(x + b)$.
${x^2} + 1$
The given polynomial is an irreducible polynomial. It has no simpler factors with integer coefficients.
${x^2} - 2$
The given polynomial is an irreducible polynomial. It has no simpler factors with integer coefficients. However, we could factor it as $(x - \sqrt 2 )\,(x + \sqrt 2 )$ but we are allowed to use irrational numbers.
Note:
Irreducibility of a polynomial depends on the coefficient numbers. To show that whether the given polynomial is irreducible or not we can use the long division method. If a polynomial with degree $2$ or higher is irreducible, then it has no real roots.
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