SOLUTION: I am in math091 at college and I am having issues solving(2x-3)^4 The only two answers I have figured out is 16x^4-81 and 16x^4+81

(2x-3)4

Here's how to do it, and the principle works for all cases of (A+B)N.
I'll go through every step in great detail.  But you will learn to shorten
the process by combining steps and doing some of the easy parts in your head:

The power is 4. 1 more than 4 is five, so write (2x) (-3) five times
with + signs between

 (2x) (-3)  +  (2x) (-3)  +  (2x) (-3)  +  (2x) (-3)  +  (2x) (-3) 


From left to right put the exponents 4,3,2,1,0 on the (2x)'s

 (2x)4(-3)  +  (2x)3(-3)  +  (2x)2(-3)  +  (2x)1(-3)  +  (2x)0(-3)

From left to right put the exponents 0,1,2,3,4 on the (-3)'s

 (2x)4(-3)0 +  (2x)3(-3)1 +  (2x)2(-3)2 +  (2x)1(-3)3 +  (2x)0(-3)4

Put a 1 coefficient before the 1st term:

1(2x)4(-3)0 +  (2x)3(-3)1 +  (2x)2(-3)2 +  (2x)1(-3)3 +  (2x)0(-3)4

Look at 1st term.  Multiply that 1 coefficient by the exponent of (2x), which is 4, 
getting 1×4 = 4 and divide that 4 by 1 since that is the 1st term, getting 4.  

Put a 4 coefficient before the 2nd term

1(2x)4(-3)0 + 4(2x)3(-3)1 +  (2x)2(-3)2 +  (2x)1(-3)3 +  (2x)0(-3)4

Look at 2nd term.  Multiply that 4 coefficient by the exponent of (2x), which
is 3, getting 4×3 = 12 and divide that 12 by 2 since that is the 2nd term, getting 6.  

Put a 6 coefficient before the 3rd term

1(2x)4(-3)0 + 4(2x)3(-3)1 + 6(2x)2(-3)2 +  (2x)1(-3)3 +  (2x)0(-3)4

Look at 3rd term.  Multiply that 6 coefficient by the exponent of (2x), which
is 2, getting 6×2 = 12 and divide that 12 by 3 since that is the 3rd term, getting 4.  

Put a 4 coefficient before the 4th term

1(2x)4(-3)0 + 4(2x)3(-3)1 + 6(2x)2(-3)2 + 4(2x)1(-3)3 +  (2x)0(-3)4

Look at 4th term.  Multiply that 4 coefficient by the exponent of (2x), which
is 1, getting 4×1 = 4 and divide that 4 by 4 since that is the 4th term, getting 1.  

Put a 1 coefficient before the 5th term.

1(2x)4(-3)0 + 4(2x)3(-3)1 + 6(2x)2(-3)2 + 4(2x)1(-3)3 + 1(2x)0(-3)4

That's the answer except for simplifying:

We can erase the 1 coefficients and the 1 exponents.  We can also erase the 
factors with 0 exponents since they are 1.  Finish simplifying:

(2x)4 + 4(2x)3(-3) + 6(2x)2(-3)2 + 4(2x)(-3)3 + (-3)4

16x4 + 4(8x3)(-3) + 6(4x2)(9) +4(2x)(-27) +(-3)4

16x4 - 96x3 + 216x2 - 216x + 81   

Edwin