How do you factor given that #f(-2)=0# and #f(x)=x^3-5x^2-2x+24#?

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Answer 1 · 2016-09-20T20:20:08

#f(x)=(x+2)(x-4)(x-3)#.

Having verified #f(-2)=0, (x+2)# is a factor of #f(x).#

Long Division is well-known Method to factorise, but we proceed as under :-

#f(x)=x^3-5x^2-2x+24#

#=ul(x^3+2x^2)-ul(7x^2-14x)+ul(12x+24)#

#=x^2(x+2)-7x(x+2)+12(x+2)#

#=(x+2)(x^2-7x+12)#

#=(x+2){ul(x^2-4x)-ul(3x+12)}#

#=(x+2){x(x-4)-3(x-4)}#

#=(x+2)(x-4)(x-3)#.

Enjoy Maths.!