Factor completely using trial and error:

$56q^3+320q^2-96q$

We have an equation$56q^3-320q^2-96q$. We will first take out the greatest common factor from the equation to reduce it to the simplest form.

Then we will find the factors accordingly and will get the factor set for the equation whose product will come out to be the given equation.

We have the equation and through it, we can see that the greatest common factor here is $8q$.

So after taking out the factor, we get a resulting equation as role="math" localid="1644576168181" $8q(7q^2-40q-12)$.

We have $8q(7q^2-40q-12)$.

We will find factors of the first and late terms.

The first term is $7$and it's factor is $(7\times 1)$.

The last term is $-12$and it's factors are role="math" localid="1644576608156" $(1\times -12),(2\times -6),(3\times -4)$.

The possible factors are-

Through this, we can see that the correct factor set is $(7q+2)(q-6)$.

Rewriting it with the greatest common factor out gives us$8q(7q+2)(q-6)$.

We will check the solution by multiplying. If we get the same given equation, our calculations are right.

So,

$$\begin{gathered} =8q(7q+2)(q-6) \\ =\left[56q^2+16q\right](q-6) \\ =56q^3-336q^2+16q^2-96q \\ =56q^3-320q^2-96q \end{gathered}$$

Thus our calculations are right.