Quick Algebra question: I went wrong somewhere.?

Question

g(x) = 3 - x^2

g(t - 1)

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heres what I did:

g(t-1) = 3 - (t-1)^2

"" = 3 - (t-1)(t-1)
"" = 3 - t^2 -t -t +1
"" = 3 - t^2 - 2t + 1
"" = -t^2 - 2t + 4
"" = t^2 +2t - 4

The answer in the book says it should be:
-t^2 +2t + 2
which would be the same as:
t^2 -2t -2

So where did I go wrong? I'm betting it has something to do with that "3 - " part.

Answer 1

Your method is pretty much correct, but just remember the order of operations and the use of signs. When you get to this point:

g(t - 1) = 3 - (t - 1)(t - 1)
g(t - 1) = 3 - (t^2 - 2t +1)

remember that the negative sign in front impacts the entire expression. Even though you are expanding it as you go along, the negative sign still applies to each component (distributive property). So if you carry on, you should get:

g(t - 1) = 3 - t^2 + 2t - 1

= -t^2 + 2t + 2 or t^2 - 2t - 2

Hope that helps!

Answer 2

I think that you had combined the 3 and 1 and then multiplied by a negative 1, where as you should have multiplied by the -1 and then took care of the 3, as in:
3-(t^2-2t+1)
3+(-t^2+2t-1)
-t^2+2t+2

Answer 3

Need to subtract the entire quantity (t-1)^2.

It's 3 - (t^2 - 2t +1)
or 3 - t^2 + 2t -1 (carry the sign change through)
or -t^2 +2t +2

Answer 4

You're right; you need to distribute that negative sign to all of (t-1)^2.

g(t-1) = 3 - (t-1)^2
= 3 - (t-1)(t-1)
= 3 - (t^2 -2t +1) Keep your parenthesis here.
= 3 - t^2 +2t -1 Distribute the negative sign here.
= - t^2 + 2t +2

Answer 5

You're good to
3 - (t-1)(t-1) But then it's
3-(t²-2t+1) distribute the - sign
3-t²+2t-1 now add them up
-t²+2t+2 Which, BTW is -not- the same as t²-2t-2.

HTH

Doug

Answer 6

Put a parethesis around the whole t-1 sq term and you get 3-1 which equals 2 ( you did 3+1)

Answer 7

thts b/c ur suppose to distribute the negative b4 u combine 3 and 1
so it would be 3-t^2+2t-1
so its -t^2+2t+2

Answer 8

when it said three minus it shoulda been comined with the 1. 3-1=2