Ashtyn Duncan

2023-03-08

What is irreducible factors?

Adriana Finley

Beginner2023-03-09Added 6 answers

It is a particular type of quadratic polynomial that cannot be divided into two linear polynomials with actual coefficients. As a result, the following cannot be written:

$F(x)=(ax+b)(cx+d)F(x)=(ax+b)(cx+d)$ where a,b,c,d

We can derive another definition.

Let's rewrite it as:

$F(x)=a(x+\frac{b}{a})c(x+\frac{d}{c})\phantom{\rule{0ex}{0ex}}=ac(x-(-\frac{b}{a}))(x-(-\frac{d}{c}))$

This polynomial is equal to 0 when x equals to -b/a or -d/c . So these numbers are the roots of this polynomial.

Because of this, we can state that the irreducible quadratic polynomial is the one without actual roots.

For example we can carry out reduction on:

${x}^{2}-5x+6=(x-2)(x-3)$

However we cannot do this with this one: ${x}^{2}+1$

$F(x)=(ax+b)(cx+d)F(x)=(ax+b)(cx+d)$ where a,b,c,d

We can derive another definition.

Let's rewrite it as:

$F(x)=a(x+\frac{b}{a})c(x+\frac{d}{c})\phantom{\rule{0ex}{0ex}}=ac(x-(-\frac{b}{a}))(x-(-\frac{d}{c}))$

This polynomial is equal to 0 when x equals to -b/a or -d/c . So these numbers are the roots of this polynomial.

Because of this, we can state that the irreducible quadratic polynomial is the one without actual roots.

For example we can carry out reduction on:

${x}^{2}-5x+6=(x-2)(x-3)$

However we cannot do this with this one: ${x}^{2}+1$