Factor this polynomial: r^4 + 4r³ + 6r² + 4r + 1?

Question

haven't learned how to factor with 5 terms... thanks in advance

Answer 1

Others seem to just give you the answer. Let me see if I can give you the way to answer. Try using synthetic (or long) division. I don't know your background in math, so it's possible that all that I say is jibberish to you. Sorry. Anyway, by using the "rational roots (or zeros) test" try to divide the polynomial by -1. 1 won't work, because of Descartes Rule of Signs. OK. Since this works, it turns out you should try in again. Well, it will work again and again. The solution is r = -1 (with a multiplicity of 4). so you have:

r = -1 r = -1 r = -1 r = -1

Moving the -1 to the left:

r + 1 = 0 r + 1 = 0 r + 1 = 0 r + 1 = 0 r + 1 = 0

Make them factors:

(r + 1)(r + 1)(r + 1)(r + 1)

and simplify:

(r + 1)^4

DONE!

I hope this helps!

--The Video Math Tutor

Answer 2

This is a binomial expansion

r^4 + 4r³ + 6r² + 4r + 1=(r+1)^4

Answer 3

r^4 + 4r³ + 6r² + 4r + 1
= (r+1)^4 by the binomial theorem

Answer 4

it's (r + 1)^4