Factor x^3 + 3x^2 + 7x + 5 completely, and find all its roots.

Step-by-step solution

1. 1

   List the candidate rational roots. The constant term is $5$ and the leading coefficient is $1$, so by the Rational Root Theorem the candidates are

   $x=\pm1,\ \pm5$

2. 2

   Test them, starting with the small ones.

   $f(-1)=-1+3-7+5=0\ \checkmark$

   So $x=-1$ is a root and $(x+1)$ is a factor. (For completeness: $f(1)=16$, $f(5)=240$, $f(-5)=-80$, so $-1$ is the only rational root.) Trying the negative candidates first pays off whenever all coefficients are positive, since a positive $x$ can never make the sum zero.

3. 3

   Divide by (x + 1) with synthetic division. Using the root $-1$ on the coefficients $1,\ 3,\ 7,\ 5$:

   $1\ \to\ 3-1=2\ \to\ 7-2=5\ \to\ 5-5=0$

   The zero remainder confirms the root, and the quotient is $x^2+2x+5$:

   $x^3+3x^2+7x+5=(x+1)\left(x^2+2x+5\right)$

4. 4

   Check whether the quadratic factors further. Its discriminant is

   $2^2-4(1)(5)=4-20=-16<0$

   Negative, so $x^2+2x+5$ is irreducible over the reals and the factoring is complete over both $\mathbb{Q}$ and $\mathbb{R}$.

5. 5

   Find all three roots. From the quadratic formula, $x=\dfrac{-2\pm\sqrt{-16}}{2}=\dfrac{-2\pm4i}{2}=-1\pm2i$. So

   $x=-1,\qquad x=-1+2i,\qquad x=-1-2i$

   Interestingly all three share real part $-1$. Completing the square explains it: $x^2+2x+5=(x+1)^2+4$, so the complex pair sits directly above and below the real root.

6. 6

   Verify. The factorisation matches the original at every integer from $-30$ to $29$ ✓; substituting $x=-1$ gives exactly $0$ ✓; and the roots sum to $-3=-b/a$ ✓ with product $-\left((-1)^2+2^2\right)=-5=-d/a$ ✓.