Faster way to factorise a cubic equation?

I am part of the way through a binomial expansion problem, and it isn't the problem itself that concerns me, what I am curious about is how the solution has factorised the equation.

At some point in the problem, I ended up with the equation

$4a^3-9a^2+5=0$ (Is this a cubic equation? I can see that it is raised to the power 3 but only has 3 terms, whereas a cubic usually has four terms. Should I be imagining the equation as $4a^3-9a^2+0a+5$? Or is this not a cubic?)

Anyhow, the solution shows this equation factorised in a single step. I have attached a screenshot of this.

Now, I know I can obtain any factors of a cubic function using the factor theorem and algebraic division, but the solution has no mention of this and seems to split the equation into a quadratic and a linear factor instantly in a single step. This has made me curious to know if there is a technique I can apply to factorise equations such as this instantly, without having to go through the long-winded factor theorem followed by algebraic division. Is this possible, and is it something I am expected to do in the context of A-Level Maths?

I am part of the way through a binomial expansion problem, and it isn't the problem itself that concerns me, what I am curious about is how the solution has factorised the equation.

At some point in the problem, I ended up with the equation

$4a^3-9a^2+5=0$ (Is this a cubic equation? I can see that it is raised to the power 3 but only has 3 terms, whereas a cubic usually has four terms. Should I be imagining the equation as $4a^3-9a^2+0a+5$? Or is this not a cubic?)

Anyhow, the solution shows this equation factorised in a single step. I have attached a screenshot of this.

Now, I know I can obtain any factors of a cubic function using the factor theorem and algebraic division, but the solution has no mention of this and seems to split the equation into a quadratic and a linear factor instantly in a single step. This has made me curious to know if there is a technique I can apply to factorise equations such as this instantly, without having to go through the long-winded factor theorem followed by algebraic division. Is this possible, and is it something I am expected to do in the context of A-Level Maths?

...In this case, we notice that the sum of the coefficients is 0, so a=1 must be a root (it doesn't effect the sum of the cubic), hence (a-1) is a factor.

Re. the division:
It's possible to write down the first and third terms of the quadratic straight off, just by considering the product creating the a^3 and constant terms. Probably easiest then to perform the full multiplication, equate terms, and second term of the qudratic comes out easily.