Respuesta :
[tex]f(x)=x^3+2x^2-5x-6[/tex]
Since all coefficients are integers, we can apply the rational zeros theorem.
The trailing coefficient is -6 with the following factors (possible values for p):
[tex]p\colon\pm1,\pm2,\pm3,\pm6[/tex]The leading coefficient is 1, with factors:
[tex]q=\pm1[/tex]Therefore, all the possible values of p/q are:
[tex]\frac{p}{q}\colon\pm\frac{1}{1},\pm\frac{2}{1},\pm\frac{3}{1},\pm\frac{6}{1}[/tex]Simplifying, the possible rational roots are:
[tex]\pm1,\pm2,\pm3,\pm6[/tex]Next, we have to check if they are roots of the polynomials by synthetic division, in which the remainder should be equal to 0.
0. Dividing ,f (x), by ,x−1,. Remainder = ,-8, ,+1, is ,NOT ,a root.
,1. Dividing ,f (x), by x+,1,. Remainder = 0, ,-1, ,IS ,a root.
,2. Dividing ,f (x), by x-2. Remainder = 0, ,+2, ,IS ,a root.
,3. Dividing ,f (x), by ,x+2,. Remainder = ,4, ,-2, is ,NOT ,a root.
,4. Dividing ,f (x), by ,x−3,. Remainder = 24,, ,+3, is ,NOT ,a root.
,5. Dividing ,f (x), by ,x+3,. Remainder = 0,, ,-3, IS ,a root.
,6. Dividing ,f (x), by ,x−6,. Remainder = 252,, ,+6, is ,NOT ,a root.
,7. Dividing ,f (x), by ,x+6,. Remainder = -120,, ,-6, is ,NOT ,a root.
Actual rational roots: A. -3, -1, 2; f(x) = (x + 3)(x + 1)(x - 2)
