# Is there a number "a" such that the equation below exists? If so what is the value of "a" and its limit.

##
#lim_(x->-2)(3x^2+ax+a+3)/(x^2+x-2)#

##### 1 Answer

#a = 15#

#lim_(x->-2) (3x^2+15x+18)/(x^2+x-2) = -1#

#### Explanation:

#x^2+x-2 = (x+2)(x-1)#

So the denominator contains exactly one factor

So in order that

#3x^2+ax+(a+3)" "# is divisible by#(x+2)#

Let

This is divisible by

Substituting

#f(-2) = 3(color(blue)(-2))^2+a(color(blue)(-2))+a+3#

#color(white)(f(-2)) = 12-2a+a+3#

#color(white)(f(-2)) = 15 - a#

So we require

With this value of

#f(x) = 3x^2+15x+18 = 3(x^2+5x+6) = 3(x+2)(x+3)#

#(3x^2+15x+18)/(x^2+x-2) = (3(color(red)(cancel(color(black)(x+2))))(x+3))/((color(red)(cancel(color(black)(x+2))))(x-1)) = (3(x+3))/(x-1)#

So:

#lim_(x->-2) (3x^2+15x+18)/(x^2+x-2) = lim_(x->-2) (3(x+3))/(x-1) = (3(color(blue)(-2)+3))/(color(blue)(-2)-1) = 3/(-3) = -1#