Question

Which is the completely factored form of 12x3 – 60x2 + 4x – 20?

Answer 1

Answer:

$4(x-5)(3{x}^{2}+1)$

Step-by-step explanation:

1) Find the Greatest Common Factor (GCF).

1 - What is the largest number that divides evenly into $12x^3,-60x^2,4x,$ and $-20$ ?

It is $4.$

2 - What is the highest degree of $x$ that divides evenly into $12x^3,-60x^2,4x,$ and $-20$ ?

It is 1, since $x$ is not in every term.

3 - Multiplying the results above,

The GCF is 4.

2)  Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)

$4(\frac{12{x}^{3}}{4}+\frac{-60{x}^{2}}{4}+\frac{4x}{4}-\frac{20}{4})$

3) Simplify each term in parentheses.

$4(3{x}^{3}-15{x}^{2}+x-5)$

4) Factor out common terms in the first two terms, then in the last two terms.

$4(3{x}^{2}(x-5)+(x-5))$

5)  Factor out the common term $x-5$.

$4(x-5)(3{x}^{2}+1)$