<PRE><B>Question 1195206</B>
&lt;br&gt;
You have already received four responses, all showing different ways to solve the problem.&lt;br&gt;
Here is yet another way that might teach you something useful.&lt;br&gt;
In a cubic equation with leading coefficient 1,&lt;br&gt;
{{{x^3+bx^2+cx+d=0}}}&lt;br&gt;
with roots p, q, and r, the following are the relationships between the roots and the coefficients of the equation:&lt;br&gt;
(1) b = -(p+q+r)
(2) c = (pq+pr+qr)
(3) d = -(pqr)&lt;br&gt;
Note it is easy to see these relationships by writing the polynomial equation in factored form and expanding:&lt;br&gt;
{{{(x-p)(x-q)(x-r)=x^3+(-p-q-r)x^2+(pq+pr+qr)x-pqr}}}&lt;br&gt;
To use these facts to solve this problem....&lt;br&gt;
Since x+2 and x-3 are factors of the polynomial, two of the roots are p=-2 and q=3.&lt;br&gt;
The coefficient of the x^2 term is 5, so from (1) above
{{{-(p+q+r)=5}}}
{{{-((-2)+(3)+r)=5}}}
{{{-(1+r)=5}}}
{{{1+r=-5}}}
{{{r=-6}}}&lt;br&gt;
So the third root is r=-6.&lt;br&gt;
The problem asks for the coefficient of the x (linear) term.  From (2) above, the coefficient of the linear term is&lt;br&gt;
{{{a=pq+pr+qr=(-2)(3)+(-2)(-6)+(3)(-6)=-6+12-18=-12}}}&lt;br&gt;
ANSWER: a = -12&lt;br&gt;
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