SOLUTION: I have tried everything for this problem and I cannot get the answer. The directions say to expand it. {{{ (t+2)(t^2+4t-3) }}}

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Question 197424: I have tried everything for this problem and I cannot get the answer.
The directions say to expand it.

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
I have tried everything for this problem and I cannot get the answer.
The directions say to expand it.
 
I have tried everything for this problem and I cannot get the answer.
The directions say to expand it.

 (t+2)(t2+4t-3)

There are 2 terms in the left set of parentheses,
and 3 terms in the right set of parentheses.  Therefore
there are 2 x 3 or 6 multiplications we must make to
remove both parentheses:

Multiply the t in the left set of parentheses by the
the t2 in the right set of parentheses, getting
t3, so we write that down on the line below it:

 (t+2)(t2+4t-3)

  t3

Multiply the t in the left set of parentheses by the
the +4t in the right set of parentheses, getting
+4t2, so we write that down next:

 (t+2)(t2+4t-3)

  t3+4t2

Multiply the t in the left set of parentheses by the
the -3 in the right set of parentheses, getting
-3t, so we write that down next:  

 (t+2)(t2+4t-3)

  t3+4t2-3t

---

Multiply the +2 in the left set of parentheses by the
the t2 in the right set of parentheses, getting
+2t2, so we write that:

 (t+2)(t2+4t-3)

  t3+4t2-3t+2t2 

Multiply the +2 in the left set of parentheses by the
the +4t in the right set of parentheses, getting
+8t, so we write that down next:

 (t+2)(t2+4t-3)

  t3+4t2-3t+2t2+8t

Multiply the +2 in the left set of parentheses by the
the -3 in the right set of parentheses, getting
-6, so we write that down next:  

 (t+2)(t2+4t-3)

  t3+4t2-3t+2t2+8t-6

At this stage, before combining like terms, we should 
always count the terms to make sure we have the same
number of terms as the number of terms in the set of 
parentheses on the left times the number of terms in 
the set of parentheses on the right, which in this
case is 2x3 or 6. And we do have 6 terms before we
combine any of them.

Finally we combine all like terms:

  t3+4t2-3t+2t2+8t-6

We combine the +4t2 and the +2t2
to get +6t2.  Also we combine the -3t 
and the +8t to get +t.  So we end up with:

  t3+6t2+5t-6

That's it!  Notice how the red expressions below form 
a pattern of the 6 multiplications of every term in the 
left set of parentheses times every term in the right 
set of parentheses:

1. (t+2)(t2+4t-3)
2. (t+2)(t2+4t-3)
3. (t+2)(t2+4t-3)
4. (t+2)(t2+4t-3)
5. (t+2)(t2+4t-3)
6. (t+2)(t2+4t-3)

Edwin

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