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Let $f(x) = x^3 + x^2 + 2x + 4$ and $g(x) = x^2 + 3x + 1$. Verify that $f(x)$ has no rational root. Can we change the constant term of $g(x)$ to a suitable complex number so that $g(x)$ now divides $f(x)$?

Using the Rational Root Theorem, the possible rational roots of $f(x)$ are: $\pm1,\pm2\pm4$. None of these numbers are roots. I don't know how to tackle the second question: it has to need a trick to solve, since the roots of $f(x)$ were a mess (I checked them online). I suppose this can be solved without factorizing $f(x)$. Any ideas?

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    You can reduce the $f(x)$ Modulo $x^2 + 3 x + p$, this yields a linear expression in $x$ with coefficients that depend on $p$. The question is then if you can find a $p$ for which this linear expression becomes identical to zero, so two equations need to be satisfied. – 2017-02-21

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Suppose such a number existed.

$g(x) = (x-z_1)(x-z_2)$ and $f(x) = (x-z_1)(x-z_2)(x-z_3)$

$-(z_1+z_2) = 3\\ -(z_1 + z_2 + z_3) = 1\\ z_3 = 2$

And you have already determined that $2$ is not a root of $f(x)$

Alternatively you could do long division.

$\frac {f(x)}{x^2 + 3x + a} = \frac{(x-2)(x^2 + 3x + a) + (8-a) x + (4+2a)}{x^2 + 3x + a}$

if $(x^2 + 3x + a)$ were to divide $f(x)$ then:

$8-a = 0$ and $4 +2a = 0$

And no such $a$ exists.