2. Factor out the GCF: r^3+〖3r〗^2+4r
r(r^2+3r+4)
r^2 (r+3r+4)
r^2 (r^2+3r+4)
r(r^3+3r^2+4r)

Question

2. Factor out the GCF: r^3+〖3r〗^2+4r 
r(r^2+3r+4) 
r^2 (r+3r+4) 
r^2 (r^2+3r+4) 
r(r^3+3r^2+4r)

rizags · Answer

what is that weird parentheses u got around the 3r?

anonymous · Answer

I don't sorry. It's for my "samplework" but I don't understand it.

rizags · Answer

ok, well to begin, your first step is correct, but the rest are not

rizags · Answer

see if you can factor the r^2+3r+4 into 2 binomials. tell me if you need help

anonymous · Answer

Can you explain to me why the first one is right?

rizags · Answer

the first one is right because in the original equation, every single term has an "r". Therefore, you can factor an r out of the entire thing

rizags · Answer

in essesnce, the GCF of that is "r", but you can factor it down smaller, namely to $$r(r+1)(r+3)$$

anonymous · Answer

Ohhh okay thank you. Can you help me with another one?

rizags · Answer

yea post it in a new question and tag me. Also please medal and fan!

anonymous · Answer

I will can I just comment it here?

rizags · Answer

fine

anonymous · Answer

3. Factor out the GCF: 〖2c〗^4+〖16c〗^3-〖8c〗^2 
c^2 (〖2c〗^2+16c-8) 
2c(c^3+〖8c〗^2-4) 
〖2c〗^2 (c^2+8c-4) 
2c(c^3+8c^2-4c)

rizags · Answer

factor out c^2 to get $$2c ^{2}(c ^{2}+8c-4)$$