Factor x^3 + 4x^2 - 5 over the rationals, then over the reals.

Step-by-step solution

1. 1

   Search for a rational root. The constant is $-5$ and the leading coefficient is $1$, so the candidates are $\pm 1,\pm 5$. Testing $x=1$:

   $1+4-5=0$

   so $(x-1)$ is a factor.

2. 2

   Divide out $(x-1)$. Writing the cubic with its missing linear term as $x^3+4x^2+0x-5$ and dividing:

   $x^3+4x^2-5=(x-1)\left(x^2+5x+5\right)$

3. 3

   Test the quadratic factor for rational roots. Its discriminant is

   $D=5^2-4(1)(5)=25-20=5$

   $5$ is positive but not a perfect square, so $x^2+5x+5$ is irreducible over the rationals while still having two real roots. That is the exact boundary between the two answers.

4. 4

   Factor over the reals. The roots are $\tfrac{-5\pm\sqrt5}{2}$, so

   $x^2+5x+5=\left(x+\frac{5+\sqrt5}{2}\right)\left(x+\frac{5-\sqrt5}{2}\right)$

5. 5

   Verify both forms. Expanding $(x-1)(x^2+5x+5)$ gives $x^3+5x^2+5x-x^2-5x-5=x^3+4x^2-5$ $\checkmark$. For the real form, the two constants sum to $5$ and multiply to $\tfrac{25-5}{4}=5$, matching the coefficients of $x^2+5x+5$ $\checkmark$.