How do you Factor completely 2x^3+5x^2-37x-602x3+5x2−37x−60?

2 Answers
Sonnhard
Jul 15, 2018

(x-4)(x+5)(2x+3)(x−4)(x+5)(2x+3)

Explanation:

Note that x=4x=4 is a solution
2*4^3+5*4^2-37*4-60=128+80-148-60=02⋅43+5⋅42−37⋅4−60=128+80−148−60=0
and

x=-5x=−5

-2*5^3+5*25+37*5-60=-250+125+185-60=0−2⋅53+5⋅25+37⋅5−60=−250+125+185−60=0
so we can

2*x^3+5*x^2-37*x-602⋅x3+5⋅x2−37⋅x−60
divide by (x-4)(x+5)(x−4)(x+5)
and we get 2x+32x+3
so the completely factorization is given by

(x-4)(x+5)(2x+3)(x−4)(x+5)(2x+3)

Cem Sentin
Jul 15, 2018

(x-4)(x+5)(2x+3)(x−4)(x+5)(2x+3)

Explanation:

2x^3+5x^2-37x-602x3+5x2−37x−60

=2x^3-8x^2+13x^2-52x+15x-602x3−8x2+13x2−52x+15x−60

=2x^2*(x-4)+13x*(x-4)+15*(x-4)2x2⋅(x−4)+13x⋅(x−4)+15⋅(x−4)

=(x-4)*(2x^2+13x+15)(x−4)⋅(2x2+13x+15)

=(x-4)*(2x^2+10x+3x+15)(x−4)⋅(2x2+10x+3x+15)

=(x-4)((2x*(x+5)+3*(x+5))(x−4)((2x⋅(x+5)+3⋅(x+5))

=(x-4)(x+5)(2x+3)(x−4)(x+5)(2x+3)

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