We have to use the Remainder Theorem for this problem.
Given a function #p(x)#, we can divide it by another function #d(x)#, to get a quotient #q(x)# and remainder #r(x)#. In other words, we can write this as #(p(x))/(d(x))=q(x)+(r(x))/(d(x))#. Now, we can multiply both sides by #d(x)# to get #p(x)=q(x)d(x)+r(x)#. If you think about it, we can find a #p(k)# for any #k# by evaluating #q(k)d(k)+r(k)#. Since we can choose #d(k)#, we should choose it so that it evaluates to 0.
In other words, for evaluating #p(k)#, we can find a #d(k)=0# and find the remainder of #(p(k))/(d(k))#, which will be the answer.
In order to evaluate #p(-3)# then, we find a #d(-3)# that is equal to #0#. One obvious example would be #d(x)=x+3#. Now, we apply long or synthetic division (using #u# for #x#):
#(13u^2+29u+25)/(u+3)=13u-10+55/(u+3)#. The remainder, 55, is the answer.
