+0  
 
+1
173
6
avatar+2442 

Still stuck on #20

20. Don't know what to do after this: [(16a^4/12a^3)]-[(40a^2/12a^3)]+[(24a/12a^3)]

 Sep 11, 2018
edited by RainbowPanda  Sep 11, 2018
 #1
avatar+2442 
+3

Nvm I got 18 but still need help with 20

 Sep 11, 2018
 #2
avatar+3994 
+3

18. \(4(2-3w)(w^2-2w+10)\), we take a good look at the last two terms, excluding the 4: (2-3w) and (w^2-2w+10)

We can expand or distribute the first two terms(2 and -3w) with the rest of the terms. So, we yield: \(2w^2+2\left(-2w\right)+2\cdot \:10+\left(-3w\right)w^2+\left(-3w\right)\left(-2w\right)+\left(-3w\right)\cdot \:10\) . We simplify this, to get: \(-3w^3+8w^2-34w+20\).

 

Now, finally, we multiply this terms by 4: \(\:4\left(-3w^3+8w^2-34w+20\right)= \boxed{-12w^3+32w^2-136w+80}\), and that's our answer!

 

I have to go now, bye!

Oops, didn't see your reply, sorry.

 

smileysmiley

 Sep 11, 2018
edited by tertre  Sep 11, 2018
 #3
avatar+2442 
+2

Oh darn >.< I just needed 20. It's alright 

RainbowPanda  Sep 11, 2018
 #4
avatar+3994 
+3

Wait, I can try 20!

20. \(\frac{16a^4-40a^2+24a}{12a^3}\), we can factor out a common term 8a from the numerator, so we get: \(8a\left(2a^3-5a+3\right)\).

Now, we are left with \(\frac{8a\left(2a^3-5a+3\right)}{12a^3}\) . We can factor a common term 4, and cancel \(a\) , a common factor. 

Cancel common factor 4: \(\frac{2a\left(2a^3-5a+3\right)}{3a^3}\).

Cancel common factor \(a\)\(\frac{2\left(2a^3-5a+3\right)}{3a^2}\).

 

So, \(\boxed{\frac{2\left(2a^3-5a+3\right)}{3a^2}} \)

is our answer!

smileysmiley

 Sep 11, 2018
 #5
avatar+3994 
+3

Notice: In the cancellation of a common factor a, we canceled 2a and 3a^3, yielding us with: 2 and 3a^2.

smileysmiley

tertre  Sep 11, 2018
 #6
avatar+2442 
+1

Thank you so much! ^-^ have an amazing day!

RainbowPanda  Sep 11, 2018

28 Online Users

avatar
avatar
avatar