Question

type the correct answer in the box. what is the completely factored form of this polynomial? $18x^3 - 120x^2 - 42x$ write the greatest common factor (gcf) first, and write all factors in standard form.

Explanation:

Step1: Find the GCF of the terms

The terms are \(18x^3\), \(-120x^2\), and \(-42x\). The GCF of the coefficients \(18\), \(-120\), and \(-42\) is \(6\), and the GCF of the variables \(x^3\), \(x^2\), and \(x\) is \(x\). So the GCF is \(6x\).
Divide each term by \(6x\):
\(\frac{18x^3}{6x}=3x^2\), \(\frac{-120x^2}{6x}=-20x\), \(\frac{-42x}{6x}=-7\)
So, \(18x^3 - 120x^2 - 42x = 6x(3x^2 - 20x - 7)\)

Step2: Factor the quadratic

Now factor the quadratic \(3x^2 - 20x - 7\). We need two numbers that multiply to \(3\times(-7)=-21\) and add to \(-20\). The numbers are \(-21\) and \(1\).
Rewrite the middle term: \(3x^2 - 21x + x - 7\)
Group the terms: \((3x^2 - 21x) + (x - 7)\)
Factor out the GCF from each group: \(3x(x - 7) + 1(x - 7)\)
Factor out \((x - 7)\): \((3x + 1)(x - 7)\)

Step3: Combine the factors

Putting it all together, the completely factored form is \(6x(3x + 1)(x - 7)\)

Answer:

\(6x(3x + 1)(x - 7)\)