Question 1007663

If       {{{f(x) = x^3-4x^2+11x-9}}}      and            {{{g(x) = x^2-4x-1}}}


  find {{{f(x)-g(x-1)}}}....since {{{f(x)= x^3-4x^2+11x-9}}} we need {{{g(x-1)=(x-1)^2-4(x-1)-1}}}

then we have:
 
{{{f(x)-g(x-1)= x^3-4x^2+11x-9-(x^2-2x+1-4x+4-1)}}}

{{{f(x)-g(x-1)= x^3-4x^2+11x-9-(x^2-6x+4)}}}

{{{f(x)-g(x-1)= x^3-4x^2+11x-9-x^2+6x-4}}}

{{{f(x)-g(x-1)= x^3-5x^2+17x-13}}}

{{{f(x)-g(x-1)= x^3-x^2-4x^2+4x+13x-13}}}

{{{f(x)-g(x-1)= (x^3-x^2)-(4x^2-4x)+(13x-13)}}}

{{{f(x)-g(x-1)= x^2(x-1)-4x(x-1)+13(x-1)}}}

{{{f(x)-g(x-1)=(x-1) (x^2-4x+13)}}}

zeros:

{{{(x-1) (x^2-4x+13)=0}}}

if {{{(x-1)=0}}}=>{{{highlight(x=1)}}} real zero 

{{{(x^2-4x+13)=0}}}...since cannot be factored, use quadratic formula

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

{{{x = (-(-4) +- sqrt( (-4)^2-4*1*13 ))/(2*1) }}}

{{{x = (4 +- sqrt( 16-52 ))/2 }}}

{{{x = (4 +- sqrt( -36 ))/2 }}}

{{{x = (4 +- 6i )/2 }}}

{{{x = (cross(4)2 +- cross(6)3i )/cross(2) }}}

{{{x = (2 +- 3i ) }}}

complex zeros:

{{{highlight(x = 2 + 3i)  }}}

{{{highlight(x = 2 - 3i ) }}}