Question

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

Answer 1

`(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!

Answer 2

`(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)`