Question

question
factor completely:
$27x^2 - 60x - 3x^3$
answer attempt 1 out of 2

Explanation:

Step1: Rearrange the polynomial

First, we rearrange the given polynomial in descending order of the exponents of \(x\). So, \(27x^{2}-60x - 3x^{3}=- 3x^{3}+27x^{2}-60x\)

Step2: Factor out the greatest common factor (GCF)

The GCF of \(-3x^{3}\), \(27x^{2}\) and \(-60x\) is \(-3x\). Factoring out \(-3x\) from each term:

$$
LATEXBLOCK0
$$

Step3: Factor the quadratic trinomial

Now, we factor the quadratic trinomial \(x^{2}-9x + 20\). We need two numbers that multiply to \(20\) and add up to \(- 9\) (but since the middle term is \(-9x\) and the constant term is \(20\), the numbers are \(-4\) and \(-5\) because \((-4)\times(-5) = 20\) and \((-4)+(-5)=-9\)). So,

$$ x^{2}-9x + 20=(x - 4)(x - 5) $$

Step4: Write the completely factored form

Substituting the factored form of the quadratic back into the expression, we get:

$$ -3x(x - 4)(x - 5)=3x(4 - x)(5 - x) \quad (\text{or we can also write it as }-3x(x - 4)(x - 5)) $$

Answer:

\(-3x(x - 4)(x - 5)\) (or \(3x(4 - x)(5 - x)\))