Question 1090160
<font color="black" face="times" size="3">A. The goal here is to first evaluate g(3). Then multiply both sides by 3 to compute 3*g(3).


g(x) = x^2 + 2x + 1


g(3) = (3)^2 + 2(3) + 1 ... replace every x with 3. Use PEMDAS to evaluate.


g(3) = 9 + 2(3) + 1


g(3) = 9 + 6 + 1


g(3) = 16


3*g(3) = 3*16 ... multiply both sides by 3


3*g(3) = <font color=red>48</font>


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B. Start by computing g(-2). Then subtract 5 from both sides


g(x) = x^2 + 2x + 1


g(-2) = (-2)^2 + 2(-2) + 1 ... every x has been replaced with -2


g(-2) = 4 + 2(-2) + 1


g(-2) = 4 - 4 + 1


g(-2) = 1


g(-2)-5 = 1-5 ... subtract 5 from both sides


g(-2)-5 = <font color=red>-4</font>


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C. This is a repeat of part B. Possibly a typo?


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D. Unlike the other parts (A and B), we aren't dealing with a single number. Instead we have an algebraic expression. The rules will effectively be the same though. 


g(x) = x^2 + 2x + 1


g(b-1) = (b-1)^2 + 2(b-1) + 1 ... replace every x with "b-1"


g(b-1) = b^2 - 2b + 1 + 2(b-1) + 1 ... FOIL


g(b-1) = b^2 - 2b + 1 + 2b - 2 + 1 ... Distribute


g(b-1) = <font color=red>b^2</font></font>