How do you factor completely #8u^2 (u+1) + 2u (u+1) -3 (u+1)#?

2 Answers
Jun 11, 2018

Answer:

#(u+1)(4u+3)(2u-1)#

Explanation:

The key realization here is that every term has a #u+1# in common, so we can factor that our. We're left with:

#(u+1)color(darkblue)((8u^2+2u-3))#

What I have in blue, we can factor by grouping. Here, I'll split the #b# term up so that we have two expressions. Here's what I mean:

#(u+1)color(darkblue)((8u^2ul(-4u+6u)-3))#

What I have underlined is the same as #2u#, so I didn't change the value of this expression.

#(u+1)(color(blue)(8u^2-4u)+color(red)(6u-3))#

I can factor a #4u# out of the blue term, and a #3# out of the red term. We now have:

#(u+1)*color(blue)(4u ul((2u-1)))*color(red)(3 ul((2u-1)))#

Both the underlined terms have a #2u-1# in common, so we can factor that out. We get:

#(u+1)(4u+3)(2u-1)#

Hope this helps!

Jun 11, 2018

Answer:

#(u+1)(2u-1)(4u+3)#

Explanation:

#"take out the "color(blue)"common factor "(u+1)#

#=(u+1)(8u^2+2u-3)#

#"factor the quadratic using the a-c method"#

#"the factors of the product "8xx-3=-24#

#"which sum to + 2 are - 4 and + 6"#

#"split the middle term using these factors"#

#8u^2-4u+6u-3larrcolor(blue)"factor by grouping"#

#=color(red)(4u)(2u-1)color(red)(+3)(2u-1)#

#"take out the "color(blue)"common factor "(2u-1)#

#=(2u-1)(color(red)(4u+3))#

#"putting it together"#

#=(u+1)(2u-1)(4u+3)#