Question 661407
{{{(-a^2+3a-2)(a^2+3a-2) }}} ...set it equal to zero and solve for {{{a}}} 


{{{(-a^2+3a-2)(a^2+3a-2)=0 }}}...here you have the product of two polynomial, and that product will e equal to zero if either one or both of them is equal to zero

so, take this factor {{{(-a^2+3a-2)=0 }}} and find {{{a}}} that makes it equal to zero

{{{(-a^2+3a-2)=0 }}}...use quadratic formula..(don't be confused with unknown variable labeled with {{{a}}}, same as a coefficient {{{-1}}})  

{{{a = (-3 +- sqrt( 3^2-4*(-1)*(-2) ))/(2*(-1)) }}}


{{{a = (-3 +- sqrt( 9-8 ))/-2 }}}

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

{{{a = (-3 +- 1)/-2 }}}

solutions:

{{{a = (-3 + 1)/-2 }}}

{{{a = -2/-2 }}}

{{{highlight(a = 1) }}}


{{{a = (-3 -1)/-2 }}}

{{{a = -4/-2 }}}

{{{highlight(a = 2) }}}


now, do the same with second factor in given product


{{{a^2+3a-2=0 }}}


{{{a = (-3 +- sqrt( 3^2-4*1*(-2) ))/(2*1)) }}}


{{{a = (-3 +- sqrt(9+8 ))/2 }}}


{{{a = (-3 +- sqrt(17 ))/2 }}}


{{{a = (-3 +- 4.12))/2 }}}


solutions:


{{{a = (-3 +4.12)/2 }}}


{{{a = 1.12/2 }}}

{{{highlight(a = 0.56) }}}


{{{a = (-3 -4.12)/2 }}}


{{{a = -7.12/2 }}}


{{{highlight(a = -3.56) }}}


so, there are four solutions because, if you multiply given polynomials you will get polynomial of fourth degree {{{-a^4+9a^2-12a+4}}}


{{{ graph(600, 600, -5, 5, -5, 55, -x^4+9x^2-12x+4) }}}