How do you find the zeroes of # f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25#?

1 Answer
Nov 27, 2015

Answer:

Factor by grouping twice, then use the difference of squares identity twice and hence find zeros:

#+-sqrt(10)/2#, #+-sqrt(5)#, #-1/5#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Factor by grouping:

#f(x) = 10x^5+2x^4-75x^3-15x^2+125x+25#

#= (10x^5+2x^4)-(75x^3+15x^2)+(125x+25)#

#= 2x^4(5x+1)-15x^2(5x+1)+25(5x+1)#

#= (2x^4-15x^2+25)(5x+1)#

#= (2x^4-10x^2-5x^2+25)(5x+1)#

#= ((2x^4-10x^2)-(5x^2-25))(5x+1)#

#= (2x^2(x^2-5)-5(x^2-5))(5x+1)#

#= (2x^2-5)(x^2-5)(5x+1)#

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

#= (sqrt(2)x-sqrt(5))(sqrt(2)x+sqrt(5))(x-sqrt(5))(x+sqrt(5))(5x+1)#

Hence #f(x)# has zeros for the following values of #x#

#+-sqrt(5)/sqrt(2) = +-sqrt(10)/2#

#+-sqrt(5)#

#-1/5#