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factor completely. $36j^{5} - 18j^{4} - 6j^{3} + 3j^{2}$

Question

factor completely.
$36j^{5} - 18j^{4} - 6j^{3} + 3j^{2}$

Explanation:

Step1: Find the GCF of coefficients and variables

First, find the greatest common factor (GCF) of the coefficients \(36\), \(-18\), \(-6\), and \(3\). The GCF of these numbers is \(3\). For the variable part, the lowest power of \(j\) is \(j^{2}\). So the GCF of the terms is \(3j^{2}\).

Step2: Factor out the GCF

Factor out \(3j^{2}\) from each term:
\[

$$\begin{align*} &36j^{5}-18j^{4}-6j^{3}+3j^{2}\\ =&3j^{2}(12j^{3}-6j^{2}-2j + 1) \end{align*}$$

\]
We can check if the polynomial inside the parentheses can be factored further. Let's try factoring by grouping. Group the first two terms and the last two terms:
\[

$$\begin{align*} &12j^{3}-6j^{2}-2j + 1\\ =&6j^{2}(2j - 1)-1(2j - 1)\\ =&(6j^{2}-1)(2j - 1) \end{align*}$$

\]
Now, put it all together:
\[
36j^{5}-18j^{4}-6j^{3}+3j^{2}=3j^{2}(6j^{2}-1)(2j - 1)
\]

Answer:

\(3j^{2}(6j^{2}-1)(2j - 1)\)