Can #y=x^3-3x^2-10x # be factored? If so what are the factors ?

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Answer 1 · 2017-07-11T02:18:50

#y=x(x-5)(x+2)#

Take x out as a factor first to give:

#y=x(x^2-3x-10)#

Then factorise the quadratic in the brackets to give the answer above.

Answer 2 · 2017-07-11T03:09:57

#y=x(x-5)(x+2)#

Inspection reveals that all three terms have #x# common in the given polynomial
#y=x^3-3x^2-10x# .....(1)

LHS can be written as
#x(x^2-3x-10)#

The quadratic in the brackets can be factorized by splitting the middle term
#(x^2-3x-10)#
#=>(x^2-5x+2x-10)#
#=>(x[x-5]+2[x-5])#
#=>([x-5][x+2])#

Hence factors are

#y=x(x-5)(x+2)#

Answer 3 · 2017-07-12T12:00:33

#y = x(x-5)(x+2)#

Take out the common factor of #x# first:

#y = x^3 -3x^2 -10x#

#y = x(x^2-3x-10)#

To factorise the quadratic trinomial, find factors of #10# which subtract to give #3#

#15xx2 = 10 and 5-2 =3# so these are the factors we need.

Their signs must be different, but there must be more negatives.
This gives:

#y = x(x-5)(x+2)#