What are all the possible rational zeros for #f(x)=5x^3+29x^2+19x-5# and how do you find all zeros?

1 Answer
Oct 24, 2016

Answer:

All the rational zeros arefor #x=-1# , #x=-5# and #x=1/5#

Explanation:

By trial and error, we can see that

#f(-1)=-5+29-19-5=0#
So #(x+1)# is a factor
So we do a long division to find the other factors
#5x^3+29x^2+19x-5##color(white)(aaaaaa)##∣x+1#
#5x^3+5x^2##color(white)(aaaaaaaaaaaaaaa)##∣##5x^2+24x-5#
#color(white)(a)# #0+24x^2#
#color(white)(aaaaa)# #24x^2+19x#
#color(white)(aaaaaaa)# #0+24x#
#color(white)(aaaaaaaaaaa)# #-5x-5#
#color(white)(aaaaaaaaaaa)# #-5x-5#
#color(white)(aaaaaaaaaaaaaa)# #0+0#
Now we have to factorize
#5x^2+24x-5=(5x-1)(x+5)#

So all the rational zeros arefor #x=-1# , #x=-5# and #x=1/5#