How do you factor #-x ^ { 3} + 8x ^ { 2} + 44x - 240= 0#?

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Answer 1 · 2017-12-17T17:41:31

#-x^3+8x^2+44x-240=-(x+6)(x-4)(x-10)#

#-x^3+8x^2+44x-240=0#

#x^3-8x^2-44x+240=0#

#x^3-4x^2-4x^2+16x-60x+240=0#

#x^2*(x-4)-4x*(x-4)-60*(x-4)=0#

#(x-4)*(x^2-4x-60)=0#

#(x-4)(x+6)(x-10)=0#

Hence #-x^3+8x^2+44x-240=-(x+6)(x-4)(x-10)#