How do you Factor completely $2 x^{3} + 5 x^{2} - 37 x - 60$ $2x^3+5x^2-37x-60$ ?

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How do you Factor completely $2 x^{3} + 5 x^{2} - 37 x - 60$ $2x^3+5x^2-37x-60$ ?

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Answer 1 · 2018-07-15T10:34:30

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

Explanation:

Note that $x = 4$ $x=4$ is a solution
$2 \cdot 4^{3} + 5 \cdot 4^{2} - 37 \cdot 4 - 60 = 128 + 80 - 148 - 60 = 0$ $2*4^3+5*4^2-37*4-60=128+80-148-60=0$
and

$x = - 5$ $x=-5$

$- 2 \cdot 5^{3} + 5 \cdot 25 + 37 \cdot 5 - 60 = - 250 + 125 + 185 - 60 = 0$ $-2*5^3+5*25+37*5-60=-250+125+185-60=0$
so we can

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

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

Answer 2 · 2018-07-15T12:38:58

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

Explanation:

$2 x^{3} + 5 x^{2} - 37 x - 60$ $2x^3+5x^2-37x-60$

= $2 x^{3} - 8 x^{2} + 13 x^{2} - 52 x + 15 x - 60$ $2x^3-8x^2+13x^2-52x+15x-60$

= $2 x^{2} \cdot (x - 4) + 13 x \cdot (x - 4) + 15 \cdot (x - 4)$ $2x^2*(x-4)+13x*(x-4)+15*(x-4)$

= $(x - 4) \cdot (2 x^{2} + 13 x + 15)$ $(x-4)*(2x^2+13x+15)$

= $(x - 4) \cdot (2 x^{2} + 10 x + 3 x + 15)$ $(x-4)*(2x^2+10x+3x+15)$

= $(x - 4) ((2 x \cdot (x + 5) + 3 \cdot (x + 5))$ $(x-4)((2x*(x+5)+3*(x+5))$

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

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How do you Factor completely $2 x^{3} + 5 x^{2} - 37 x - 60$ $2x^3+5x^2-37x-60$ ?

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How do you Factor completely $2 x^{3} + 5 x^{2} - 37 x - 60$ $2x^3+5x^2-37x-60$ ?