What are all the possible rational zeros for #f(x)=2x^3+5x^2+4x+1# and how do you find all zeros?

1 Answer
Sep 11, 2016

Answer:

the zeroes are. #-1,-1,-1/2#.

Explanation:

#f(x)=2x^3+5x^2+4x+1#

Note that, the sum of the co-effs. of odd-powered terms of

#x#=2+4=6, and, that of even-powered#=5+1=6.#.

#:. (x+1)" is a factor of "f(x)#.

Now, #f(x)=2x^3+5x^2+4x+1#

#=ul(2x^2+2x^2)+ul(3x^2+3x)+ul(x+1)#

#=2x^2(x+1)+3x(x+1)+1(x+1)#

#=(x+1)(2x^2+3x+1)#

#(x+1){ul(2x^2+2x)+ul(x+1)}#

#=(x+1){2x(x+1)=1(x+1)}#

#=(x+1){(x+1))(2x+1)}#

#=(x+1)^2(2x+1)#

Hence, the zeroes are. #-1,-1,-1/2#.