How do you find all the zeros of #f(x)=x^3 + 4x^2 - 5#?

1 Answer

Answer:

#x_1=1# and #x_2=(-5+sqrt5)/2# and #x_3=(-5-sqrt5)/2#

Explanation:

From the given:
#y=x^3+4x^2-5#

Perform synthetic division

#x^3" " " "x^2" " " "x^1" " " " "x^0#

#1" " " " " 4" " " " "0" " " -5# trial divisor #=1#
#underline(" " " " " " "1" " " " " "5" " " " " "5)#
#1" " " " " 5" " " " " "5" " " " " " 0##larr#remainder

#x_1=1# is a zero

depressed equation:

#x^2+5x+5=0#

by quadratic formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Let #a=1# and #b=5# and #c=5#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-5+-sqrt(5^2-4(1)(5)))/(2(1))#

#x=(-5+-sqrt5)/2#

the other zeros are:
#x_2=(-5+sqrt5)/2#

#x_3=(-5-sqrt5)/2#

God bless....I hope the explanation is useful.