How do you Factor completely #2x^3+5x^2-37x-60#?

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Answer 1 · 2018-07-15T10:34:30

#(x-4)(x+5)(2x+3)#

Note that #x=4# is a solution
#2*4^3+5*4^2-37*4-60=128+80-148-60=0#
and

#x=-5#

#-2*5^3+5*25+37*5-60=-250+125+185-60=0#
so we can

#2*x^3+5*x^2-37*x-60#
divide by #(x-4)(x+5)#
and we get #2x+3#
so the completely factorization is given by

#(x-4)(x+5)(2x+3)#

Answer 2 · 2018-07-15T12:38:58

#(x-4)(x+5)(2x+3)#

#2x^3+5x^2-37x-60#

=#2x^3-8x^2+13x^2-52x+15x-60#

=#2x^2*(x-4)+13x*(x-4)+15*(x-4)#

=#(x-4)*(2x^2+13x+15)#

=#(x-4)*(2x^2+10x+3x+15)#

=#(x-4)((2x*(x+5)+3*(x+5))#

=#(x-4)(x+5)(2x+3)#