How do you find all the zeros of #f(x)=4x^3-20x^2-3x+15#?

1 Answer
Feb 27, 2016

Answer:

Factor by grouping and by using the difference of squares identity to find:

#f(x) =(2x-sqrt(3))(2x+sqrt(3))(x-5)#

hence has zeros #x = +-sqrt(3)/2# and #x = 5#

Explanation:

Factor by grouping, then use the difference of squares identity:

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

with #a=2x# and #b=sqrt(3)#, as follows:

#f(x) = 4x^3-20x^2-3x+15#

#=(4x^3-20x^2)-(3x-15)#

#=4x^2(x-5)-3(x-5)#

#=(4x^2-3)(x-5)#

#=((2x)^2-(sqrt(3))^2)(x-5)#

#=(2x-sqrt(3))(2x+sqrt(3))(x-5)#

So the zeros of #f(x)# are #x = +-sqrt(3)/2# and #x = 5#