SOLUTION: Hi! Please help me solve Given g(x)=x^2+2x+1 a.3g(3) b.g(-2)-5 c.g(-2)-5 d.g(b-1) Thank you!

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Question 1090160: Hi! Please help me solve
Given g(x)=x^2+2x+1
a.3g(3)
b.g(-2)-5
c.g(-2)-5
d.g(b-1)
Thank you!
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
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) = 48

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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 = -4

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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) = b^2