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

1 Answer
Sep 28, 2016

Answer:

The possible rational zeros are #1, -1, 5, -5#.
The zeros are #x=-5, x=2+sqrt3, x=2-sqrt3#.

Explanation:

#f(x)=color(blue)1x^3+x^2-19x+color(red)5#

The possible rational zeros are the factors of the constant term #color(red)5# divided by the factors of the leading coefficient #color(blue)1#.

#x=(+-1,+-5)/(+-1) =1, -1,5, -5#

To find all the zeros, choose one of the possible zeros and use synthetic division. If the remainder is zero, you have verified that the possible zero is in fact a zero.

Choosing #-5# as a possible zero:

#color(white)---------#
#-5]color(white)a1color(white)(aaaa)1color(white)(aa)-19color(white)(ax^111)5#
#color(white)(a^1a^1)darrcolor(white)(1)-5color(white)(a^1+)20color(white)(x^1)-5#
#color(white)---------#
#color(white)(a^1a^11)1color(white)(x^2)-4color(white)(aaaaa)1color(white)(aax^11)0#

The remainder is zero, so #-5# is confirmed as a zero.

The coefficients of the result of synthetic division give
#x^2-4x+1#. Because this is not factorable, use the quadratic formula to find the last two zeros.

#x=(4+-sqrt((-4)^2-4*1*1))/2=(4+-sqrt12)/2=(4+-2sqrt3)/2#

#x=2+-sqrt3#

The zeros are #x=-5, x=2+sqrt3, x=2-sqrt3#