given that g(x)=3x^2+3x+20, find each of the following g(4) and g(p+4)

Question

 I am needing help in finding what g(4) and g(p+4) is equivalent to. The problem gives the information stating that g(x) is equal to 3x squared plus 3x+ 20

Answer 1

g(x)=3x^2+3x+20

 

now replace x with 4 then

 

g(4)=3*(4^2)+3*4+20

 

       =3*16+12+20

       

       = 48+12+20

       = 80

        ============

 

 

Now replace x with (p+4), then

 

g(p+4)=3*(p+4)^2+3*(p+4)+20                                      (a+b)^2=a^2+2ab+b^2

 

                                                                                      (p+4)^2=p^2+2*p*4+4^2

          =3[p^2+2*p*4+4^2]+3p+3*4+20

         

          =3[p^2+8p+16]+3p+12+20

 

          =[3*p^2+3*8p+3*16]+3p+32

 

          =3*p^2+24p+48+3p+32

 

          =3*p^2+27p+80

 

           =3p²+27p+80

        ============================

 

Answer 2

Plug in 4 into x to get g(4).

g(4)=3(4)^2+3(4)+20=3(16)+12+20=80

Plug in p+4 into x to get g(p+4)

g(p+4)=3(p+4)^2+3(p+4)+20

           =3(p^2+8p+16)+3p+12+20

           =3p^2+24p+48+3p+32

            =3p^2+27p+80

 

now: g(4)+g(p+4) = 80+3p^2+27p+80

                            =3p^2+27p+160