# How does this simplify to this?

• Dec 18th 2010, 11:07 PM
Klutz
How does this simplify to this?
So I was just doing a calculus question and on the solutions ..

it jumps from

$\frac{3(x+2)^2(x^2+1)-2x(x+2)^3}{(x^2+1)^2}$

then it skips to

$\frac{(3(x^2+1)-2x(x+2))(x+2)^2}{(x^2+1)^2}$

and I don't know how that works.. there have been a number of questions that have done that and now I think its time i asked about it. Of course, thats not the simplest form but I understand how it simplifies from there.. I just don't get how it skips from the first one to the second.

If someone could care to explain I'd really love it thanks :D
• Dec 18th 2010, 11:12 PM
Prove It
Are you sure it's not $\displaystyle \frac{3(x+2)^2(x^2+1) - 2x(x+2)^3}{(x^2+1)^2}$?

If it is, they've realised there's a common factor of $\displaystyle (x+2)^2$ in the numerator.
• Dec 18th 2010, 11:22 PM
Klutz
sorry yeah i'll fix it my bad!
• Dec 18th 2010, 11:24 PM
Klutz
I still don't get it though.. could you explain in more depth?
• Dec 18th 2010, 11:33 PM
Prove It
$\displaystyle 3(x+2)^2(x^2+1) - 2x(x+2)^3 = 3(x+2)^2(x^2+1) - 2x(x+2)(x+2)^2$.

Do you see that there is a common factor of $\displaystyle (x+2)^2$?
• Dec 18th 2010, 11:40 PM
Klutz
Yeah I see it but why is it that there are 5 of them in the first equation and only 3 in the 2nd one?
• Dec 19th 2010, 12:16 AM
HallsofIvy
ab+ ac= a(b+ c) has, in your sense, "two" a's on the left and only one on the right!

6+ 10= 2(3)+ 2(5)= 2(3+ 5) has two "2"s on the left and only one on the right. They are not supposed to be the same!
• Dec 19th 2010, 12:28 AM
Klutz
I guess that makes sense.. still don't completely understand though D:
• Dec 19th 2010, 06:46 AM
Soroban
Hello, Klutz!

Quote:

$\dfrac{3(x+2)^2(x^2+1)-2x(x+2)^3}{(x^2+1)^2} \;=\;\dfrac{(x+2)^2\bigg[3(x^2+1)-2x(x+2)\bigg]}{(x^2+1)^2}$

Examine the numerator: . $3(x+2)^2(x^2+1) - 2x(x+2)^3$

The two terms have a common factor: . $3\underbrace{(x+2)^2}(x^2+1) \;-\; 2x\underbrace{(x+2)^3}$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . $_{\text{They have a common factor of }(x+2)^2}$

Factor it out: . $(x+2)^2\,\bigg[3(x^2+1) - 2x(x+2)\bigg]$

• Dec 19th 2010, 03:00 PM
Klutz
ahh! I got it thanks so much. Gosh I'm so slow.