Question 79357
The given equation is a polynomial of degree 2 and hence a quadratic equation. 


The given equation can be written and solved as follows:  



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



{{{y = x^2 + 2x + x + 2 }}}


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


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


hence, the zeroes of the polynomial is x = -2, -1 


The graph of the given quadratic is : 



{{{ graph( 400, 400, -15, 15, -15, 15, x^2 + 3x + 2 ) }}}