Question

find factors of 1+2x+3x^2+x^3 using rational root theorm

Answer 1

1 and 1+2x+3x^2+x^3

Answer 2

To find factors of the polynomial $P(x) = 1+2x+3x^2+x^3$ using the Rational Root Theorem, we first write the polynomial in standard form:

$P(x) = x^3 + 3x^2 + 2x + 1$.

The Rational Root Theorem states that if a polynomial $a_nx^n + \dots + a_1x + a_0$ has integer coefficients, then any rational root $p/q$ (where $p$ and $q$ are integers with no common factors other than 1) must satisfy:

1. $p$ is a factor of the constant term $a_0$.
2. $q$ is a factor of the leading coefficient $a_n$.

For our polynomial $P(x) = x^3 + 3x^2 + 2x + 1$:

• The constant term $a_0 = 1$. The factors of $a_0$ are $p \in \{\pm 1\}$.
• The leading coefficient $a_n = 1$. The factors of $a_n$ are $q \in \{\pm 1\}$.

Therefore, the possible rational roots $p/q$ are $\frac{\pm 1}{\pm 1}$, which simplifies to $\{\pm 1\}$.

Now, we test each of these possible rational roots:

1. Test $x = 1$:
   $P(1) = (1)^3 + 3(1)^2 + 2(1) + 1 = 1 + 3(1) + 2(1) + 1 = 1 + 3 + 2 + 1 = 7$.
   Since $P(1) \neq 0$, $x=1$ is not a root of the polynomial.

2. Test $x = -1$:
   $P(-1) = (-1)^3 + 3(-1)^2 + 2(-1) + 1 = -1 + 3(1) - 2 + 1 = -1 + 3 - 2 + 1 = 1$.
   Since $P(-1) \neq 0$, $x=-1$ is not a root of the polynomial.

Since none of the possible rational roots are actual roots, the polynomial $x^3 + 3x^2 + 2x + 1$ has no rational roots.

Conclusion:
For a cubic polynomial with integer coefficients, if it has no rational roots, then it cannot be factored into linear factors with rational coefficients. Consequently, it is irreducible over the set of rational numbers ($\mathbb{Q}$). This means it cannot be expressed as a product of two or more non-constant polynomials with rational coefficients.

Therefore, the polynomial $1+2x+3x^2+x^3$ has no non-trivial factors over the rational numbers. Its only factors are $1$ and the polynomial itself.