How do you find all the zeros of #f(x)=4x^3-12x^2-x+3#?

1 Answer
George C.
Aug 9, 2016

#f(x)# has zeros: #1/2#, #-1/2#, #3#

Explanation:

Notice that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

#f(x) = 4x^3-12x^2-x+3#

#=(4x^3-12x^2)-(x-3)#

#=4x^2(x-3)-1(x-3)#

#=(4x^2-1)(x-3)#

#=((2x)^2-1^2)(x-3)#

#=(2x-1)(2x+1)(x-3)#

Hence the zeros of #f(x)# are: #1/2#, #-1/2#, #3#

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