Question

list all the possible rational and actual rational roots for the equation x^3+2x^2-5x-6=0

Answer 1

First look at the multiples of the constant in x^3+2x^2-5x-6=0,
\[[6]=\pm1,\pm2,\pm3,\pm6\]
You can try an error by substituting these\[ \pm1,\pm2,\pm3,\pm6\]
Into the equation and see whether you get zero.
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In this case, x+1 is a factor of x^3+2x^2-5x-6=0, So,

(x+1)( Quadratic )
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To find the quadratic, use x+1 and do long division,

x^2+x-6
_______________________________
x+1) x^3+2x^2-5x-6
x^3+x^2
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x^2-5x
x^2+x
--------------------
-6x-6
-6x-6
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So,
(x+1)( Quadratic )
(x+1)( x^2+x-6 )
Factor the quadratic,
(x+1)(x-2)(x+3)
Roots are x=-1,x=2,x=-3