What are all the possible rational zeros for #f(x)=x^3+5x^2-48x-10# and how do you find all zeros?

1 Answer
Oct 20, 2016

Answer:

The only rational zero is #5#.

Explanation:

As putting #x=5# in #f(x)=x^3+5x^2-48x-10#

gives #f(x)=5^3+5xx5^2-48xx5-10=125+125-240-10=0#

Hence, #5# is a zero of #f(x)# and #(x-5)# is a factor of #f(x)#

Hence #f(x)=x^3+5x^2-48x-10#

= #x^2(x-5)+10x(x-5)+2(x-5)#

= #(x-5)(x^2+10x+2)#

As the determinant in #x^2+10x+2# is #10^2-4xx1xx2=100-8=92#, which though positive, is not a square of a rational number, there are no other rational zeros.

Irrational zeros can be had using quadratic formula as #(-10+-sqrt(100-4xx1xx2))/2=(-10+-sqrt92)/2=(-10+-2sqrt23)/2#

i.e. #-5-sqrt23# and #-5+sqrt23#