You have to check which divisors of the constant term, $4$, are the roots of $-\lambda^3-3 \lambda^2+4=0$. Let $a$ be this divisor(if there are more than one, take one of them), then apply the Euclidean division of $-\lambda^3-3 \lambda^2+4$ and $\lambda-a$.
Then you get $-\lambda^3-3 \lambda^2+4=(\lambda-a) \cdot q$, where $q$ is a second degree polynomial, which you can find its roots using the disciminant.
I might be misunderstanding, but this just looks like concealed expanding $$(x^2+ax+b)(x^2+cx+d)\tag{1}$$ and comparing coefficients.
- Factoring extremes
Well, what we want is to choose $b$ and $d$ such that $(x^2+b)(x^2+d)$ matches leading and constant factor. What it amounts to is finding $b$ and $d$ such that $bd = -3$. Let's say we were lucky and chose the winning pair $b = -1,\ d=3.$
- Calculating $\Delta$
Why is this important? Well, $(x^2-1)(x^2+3) = x^4+2x^2-3$, so we are missing one $x^2$: $\Delta = 3x^2 - 2x^2$.
- Covering for the missing $\Delta$
There is only one place where $\Delta$ can come from, that is from $a$ and $c$ - we must have $ac x^2 = \Delta$. Why? Well, if you expand $(1)$, you will find that $x^2$ term must be equal to $b + d + ac$. We found $b+d$ in step 1 and 2, so $acx^2$ is equal to $\Delta$. In this case $ac = 1$. There are two possibilities in $\Bbb Z$, either $a=c=1$, or $a=c=-1$. It is easy to decide which one, since it must fit the other terms as well, namely $x^3$ term is equal to $a + c$ and the $x$ term is equal to $ad + bc = 3a - c$. Thus, $a = c = 1$.
- Collecting terms
We simply write $(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3$ and feel the magic all around us.
Now, let us do the same thing, only transparently:
$$(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$
so we need to solve the following system in $\Bbb Z$:
\begin{array}{c r}\begin{align}
a+c&=2\\
b+d+ac&=3\\
ad+bc&=2\\
bd &= -3
\end{align} &\tag{2}\end{array}
There are two possibilities, either $b = 1$ and $d = -3$, or $b = -1$ and $d = 3$. We can see that former leads to no solution, so we will pursuit the latter, the system becomes:
\begin{align}
a+c&=2\\
ac&=1\\
3a-c&=2
\end{align}
The first and the third equation form linear system, which has unique solution $a = c = 1$, luckily, consistent with the second equation. Thus, we now know for certain that $$(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3.$$
Addendum:
There is a legitimate question why we would consider $(1)$ in the first case. Now, we are trying to factor a quartic over $\Bbb Z$. Let's say $p(x) = x^4 +2x^3+3x^2+2x-3$ and assume that $p = fg$. We have that $\deg p = \deg f + \deg g$, so there are two cases: $4=1+3$ or $4=2+2$. The first case would imply that $p$ has integer root and we can use rational root theorem to rule that out. Thus, only case $4=2+2$ remains, which leads to considering $(1)$.
Finally, if the system $(2)$ had no solutions in $\Bbb Z$, we would conclude that $p$ is irreducible over $\Bbb Z$.
Best Answer
Given your quadratic $7x^2-9x+2$, you need to think of two numbers $a$ and $b$ such that $a\times b=7$, and two numbers $c$ and $d$ such that $c\times d=2$, and $ad+bc=-9$.
It takes some time and practice to start seeing what numbers work. For example, letting $a=1, b=7, c=-1, d=-2$ works because $ab=1\times 7=7$, $cd=-1\times -2=2$, and $ad+bc=1\times-2+7\times-1=-2-7=-9$.
In general for a quadratic of the form $\alpha x^2+\beta x+\gamma$, you want $ab=\alpha$, $cd=\gamma$ and $ad+bc=\beta$, where $\alpha,\beta, \gamma\in\mathbb{R}.$