Question 1132057


The equation of {{{y=x^3+4x^2+x-6}}} has roots {{{alpha}}}, {{{beta}}} and {{{gamma}}}}. 

Find the value of 

(i)  {{{alpha^2+beta^2+gamma^2}}}

(ii) {{{alpha^2*beta^2*gamma^2}}}

(iii) 
{{{(alpha-beta)(beta+gamma)(gamma+alpha)}}}


{{{y=x^3+4x^2+x-6}}}.......factor completely


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


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


{{{y=x^2(x+3)+2x(x+3)-x(x+3)-2(x+3)}}}


{{{y = (x^2+2x-x-2)(x + 3)}}}


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


solutions:

{{{x=1}}}
{{{x=-2}}}
{{{x=-3}}}

=> let's {{{alpha=1}}}, {{{beta=-2}}} and {{{gamma=-3}}}}

then,

 (i)  

{{{alpha^2+beta^2+gamma^2}}}

{{{1^2+(-2)^2+(-3)^2}}}

{{{1+4+9}}}

{{{14}}}




(ii) 

{{{alpha^2*beta^2*gamma^2}}}

{{{1^2*(-2)^2*(-3)^2}}}

{{{1*4*9}}}

{{{36}}}


(iii) 


{{{(alpha-beta)(beta+gamma)(gamma+alpha)}}}

{{{(1-(-2))(-2-3)(-3+1)}}}

{{{(1+2)(-5)(-2)}}}

{{{(3)(10)}}}

{{{30}}}