Factor. 6j^3 + 3j^2 - 16j - 8 ______

Grouping for Factoring: Group the terms into two pairs to prepare for factoring by grouping. Group the first two terms and the last two terms separately. 6j^3 + 3j^2 - 16j - 8 can be grouped as (6j^3 + 3j^2) + (-16j - 8).Factor out Common Factors: Factor out the greatest common factor from each group. From the first group (6j^3 + 3j^2), we can factor out 3j^2, which gives us 3j^2(2j + 1). From the second group (-16j - 8), we can factor out -8, which gives us -8(2j + 1). Now we have 3j^2(2j + 1) - 8(2j + 1).Factor out Binomial Factor: Factor out the common binomial factor from the two terms. Both terms have a common factor of (2j + 1), so we factor this out. This gives us (2j + 1)(3j^2 - 8).Check for Further Factoring: Check if the second factor (3j^2 - 8) can be factored further. The second factor is a difference of squares, as 8 is a perfect square (2^3). However, 3j^2 is not a perfect square, so (3j^2 - 8) cannot be factored further using real numbers.

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Algebra 2

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Q. Factor.

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6j^3 + 3j^2 - 16j - 8

Grouping for Factoring: Group the terms into two pairs to prepare for factoring by grouping. Group the first two terms and the last two terms separately. $6j^3 + 3j^2 - 16j - 8$ can be grouped as $(6j^3 + 3j^2) + (-16j - 8)$ .
Factor out Common Factors: Factor out the greatest common factor from each group. $\newline$ From the first group $6j^3 + 3j^2$ , we can factor out $3j^2$ , which gives us $3j^2(2j + 1)$ . $\newline$ From the second group $-16j - 8$ , we can factor out $-8$ , which gives us $-8(2j + 1)$ . $\newline$ Now we have $3j^2(2j + 1) - 8(2j + 1)$ .
Factor out Binomial Factor: Factor out the common binomial factor from the two terms. $\newline$ Both terms have a common factor of $2j + 1$ , so we factor this out. $\newline$ This gives us $2j + 1$ $(3j^2 - 8)$ .
Check for Further Factoring: Check if the second factor $(3j^2 - 8)$ can be factored further. $\newline$ The second factor is a difference of squares, as $8$ is a perfect square $(2^3)$ . $\newline$ However, $3j^2$ is not a perfect square, so $(3j^2 - 8)$ cannot be factored further using real numbers.