Question 995054
a. 

{{{g(x) = x^3-2x^2 -5x + 6}}}......set {{{g(x) =0}}} and factor completely

{{{0 = x^3-2x^2+x -6x + 6}}}

{{{0 = (x^3-2x^2+x) -(6x - 6)}}}

{{{0= x(x^2-2x+1) -6(x - 1)}}}

{{{0= x(x-1)^2 -6(x - 1)}}}

{{{0 = (x(x-1) -6)(x - 1)}}}

{{{0= (x^2-x -6)(x - 1)}}}

{{{0 = (x^2-3x+2x -6)(x - 1)}}}

{{{0= (x(x-3)+2(x -3))(x - 1)}}}

{{{0 = (x-3) (x-1) (x+2)}}}

solutions:

if {{{0 = (x-3) }}}=>{{{x=3}}}
if {{{0 =  (x-1) }}}=>{{{x=1}}}
if {{{0 =  (x+2)}}}=>{{{x=-2}}}

so, here you have three real zeros

see it on a graph:

{{{ graph( 600, 600, -10, 10, -10, 10, x^3-2x^2 -5x + 6) }}}



b. 

{{{m(x) = x^3 + 4x^2 + 4x + 3}}}


{{{0 = x^3 + 3x^2+x^2 + 3x+x + 3}}}

{{{0 = (x^3 + 3x^2)+(x^2 +3x)+ (x + 3)}}}

{{{0 = x^2(x + 3)+x(x +3)+ (x + 3)}}}

{{{0 = (x+3) (x^2+x+1)}}}

solutions:

if {{{0 = (x+3) }}}=>{{{x=-3}}}=> one real solution

if {{{0 = (x^2+x+1)}}}...use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-1 +- sqrt( 1^2-4*1*1 ))/(2*1) }}} 

{{{x = (-1 +- sqrt( 1-4 ))/2 }}} 

{{{x = (-1 +- sqrt( -3 ))/2 }}} 

{{{x = (-1 +- i*sqrt( 3 ))/2 }}} 


so, you have two imaginary zeros

{{{x = (-1 + i*sqrt( 3 ))/2 }}} 
and

{{{x = (-1 - i*sqrt( 3 ))/2 }}} 


{{{ graph( 600, 600, -10, 10, -10, 10, x^3 + 4x^2 + 4x + 3) }}}