Question

the function $f(x) = x^4 + 3x^3 + 9x^2 + 21x + 14$ has at least two rational roots. use the rational root theorem to find those roots, then proceed to find all complex roots. (note: roots may be integer, rational, irrational, and/or complex.)

answer attempt 1 out of 2

there is one root

Explanation:

Step1: Apply Rational Root Theorem

The Rational Root Theorem states that possible rational roots are $\pm\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. For $f(x)=x^4 + 3x^3 + 9x^2 + 21x + 14$, leading coefficient is $1$ (so $q = \pm1$) and constant term is $14$ (factors $p=\pm1,\pm2,\pm7,\pm14$). Possible roots: $\pm1,\pm2,\pm7,\pm14$.

Test $x = -1$: $f(-1)=1 - 3 + 9 - 21 + 14 = 0$. So $x = -1$ is a root.

Test $x = -2$: $f(-2)=16 - 24 + 36 - 42 + 14 = 0$. So $x = -2$ is a root.

Step2: Factor the Polynomial

Since $x = -1$ and $x = -2$ are roots, $(x + 1)(x + 2)$ are factors. Divide $f(x)$ by $(x + 1)(x + 2)=x^2 + 3x + 2$.

Using polynomial long division or synthetic division:

$f(x)=(x^2 + 3x + 2)(x^2 + 0x + 7)=x^4 + 3x^3 + 9x^2 + 21x + 14$ (check: $(x^2 + 3x + 2)(x^2 + 7)=x^4 + 7x^2 + 3x^3 + 21x + 2x^2 + 14=x^4 + 3x^3 + 9x^2 + 21x + 14$).

Step3: Find Roots of Quadratic

Solve $x^2 + 7 = 0$. $x^2 = -7 \implies x = \pm i\sqrt{7}$.

Answer:

The rational roots are $x = -1$ and $x = -2$. The complex roots (including rational) are $x = -1$, $x = -2$, $x = i\sqrt{7}$, and $x = -i\sqrt{7}$.