expand using the binomal therorem (3x+2)^4?

Answer 1

Use the fourth row of Pascal's triangle 1 4 6 4 1 for the coefficients
For the kth term, use (3x)^k(2^(4-k-1))
So your expansion looks like tihs:
1(3x)^4)*(2^0) + 4(3x)^3)(2^1) + 6((3x)^2)*(2^2) + 4((3x^1)*(2^3) + 1(3x)^0)(2^4) =
81x^4 + 216x^3 + 216x^2 + 96x + 16
Work it out yourself to make sure I got it right and make sure you can do it for yourself.
for more info see
http://www.purplemath.com/modules/binomial.htm

Answer 2

Start with the fourth row of Pascal's triangle:
Row 0: ...... 1
Row 1: ..... 1 1
Row 2: ... 1 2 1
Row 3: . 1 3 3 1
Row 4: 1 4 6 4 1 <--

Next you need to make all combinations of 3x^a and 2^b, where the exponents (a+b) add to 4.
(3x)^4 = 81x^4
(3x)^3 * 2^1 = 54x^3
(3x)^2 * 2^2 = 36x^2
(3x)^1 * 2^3 = 24x
2^4 = 16

Now just multiply the elements of row 4 of Pascal's triangle by the numbers from before:
1 * 81x^4 = 81x^4
4 * 54x^3 = 216x^3
6 * 36x^2 = 216x^2
4 * 24x = 96x
1 * 16 = 16

The final answer is:
81x^4 + 216x^3 + 216x^2 + 96x + 16

Answer 3

From Pascal's Triangle, the coefficients of the terms in a fourth degree binomial are 1, 4, 6, 4, 1, in that order. So a binomial (a + b) raised to the fourth power will look like this:

a^4 + 4a³b + 6a²b² + 4ab³ + b^4.

Now all we have to do is let a = 3x and b = 2 and put them in their appropriate places in the above polynomial. When we do that, we get:

****[(3x)²]² + 4[2(3x)³] + 6[2²(3x)²] + 4[2³(3x)] + [(2)²]² = ****

81x^4 + 216x³ + 216x² + 96x + 16, which is your answer.

Don't let the exponentiation of the first and last terms in the starred line throw you. I was simply trying to make them all look consistent. They both are the number in the brackets raised to the fourth power. Also, since b = 2 is a constant, I pulled it out in front of the variable 3x to be consistent with normal mathematical notation. A constant normally precedes a variable.

Answer 4

easy
81x^4 + 216x^3 + 216x^2 + 96x+ 16

Answer 5

(a+b)^4=a^4+4a^b+6a^2b^2+4ab^3+b^4
a=3x
b=2

(3x+2)^4=(3x)^4+4(3x)^3*2+6(3x)^2*2^2+4*(3x)*2^3+2^4
(3x+2)^4=81x^4+216x^3+216x^2+96x+16