How do you factor #y=x^3-4x^2-17x+60# ?

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Answer 1 · 2016-02-09T04:56:10

The factors are #y# =#(x-3)# * #(x-5)# * #(x+4)#

Assume that the factors are #(x-a)#, #(x-b)# and #(x-c)#. Two things arise from this.

(i) if #x# is either equal to #a# or #b# or #c#, #y=0# and

(ii) #a*b*c=-60#, which means #a# or #b# and #c# are factors of 60.

So, which factor of 60 (such as 2, 3, 5, 6, 10 etc.) makes #y=0#.

Hit and trial will show that #x=3# returns #y=0#

and hence #(x-3)# is a factor of #y#.

When we divide #y# by #(x-3)#, we get #x^2-x-20#

Similarly #x=5# makes #x^2-x-20# equal to zero

and hence next factor is #(x-5)#

Dividing #x^2-x-20# by #(x-5)#, we get #(x+4)#

Hence the factors of #y=x^3-4x^2-17x+60# are

#y# =#(x-3)# * #(x-5)# * #(x+4)#