How do you expand (x+1)^2(x-3)^2????????

[x^2 -2x -3]^2

here u go
(x+1)^2(x-3)^2
=(x^2+1+2x)(x2+9-6x)
=x^4+9x^2-6x^3+x^2+9-6x+2x^3+18x-12x^2
=x^4-2x^2-4x^3+12x+9 Answer

(x + 1)^2(x - 3)^2
= [(x + 1)(x - 3)]^2
= [x*x + 1*x - x*3 - 1*3]^2
= [x^2 + x - 3x - 3]^2
= [x^2 - 2x - 3]^2
= [x^2 - 2x - 3][x^2 - 2x - 3]
= x^2*x^2 - 2x*x^2 - 3*x^2 - x^2*2x + 2x*2x + 3*2x - x^2*3 + 2x*3 + 3*3
= x^4 - 2x^3 - 3x^2 - 2x^3 + 4x^2 + 6x - 3x^2 + 6x + 9
= x^4 - 2x^3 - 2x^3 - 3x^2 + 4x^2 - 3x^2 + 6x + 6x + 9
= x^4 - 4x^3 - 2x^2 + 12x + 9

You do have the first part right, good job.
The second part is harder. Each term in the first polynomial must be multiplied by each term in the second polynomial.

You will come up with nine terms, starting with x^4, and ending with 1*9
After you have all nine terms, group those that have the same exponent:
-6x^3+2x^3=-4x^3
for example.

good luck

(x + 1)^2(x - 3)^2
= [(x + 1)(x - 3)]^2
= [x*x + 1*x - x*3 - 1*3]^2
= [x^2 + x - 3x - 3]^2
= [x^2 - 2x - 3]^2
= [x^2 - 2x - 3][x^2 - 2x - 3]
= x^2*x^2 - 2x*x^2 - 3*x^2 - x^2*2x + 2x*2x + 3*2x - x^2*3 + 2x*3 + 3*3
= x^4 - 2x^3 - 3x^2 - 2x^3 + 4x^2 + 6x - 3x^2 + 6x + 9
= x^4 - 2x^3 - 2x^3 - 3x^2 + 4x^2 - 3x^2 + 6x + 6x + 9
= x^4 - 4x^3 - 2x^2 + 12x + 9

Edit: Σducator, please don't copy my answer.
Edit(1): Henri C, please don't copy my answer.

(x+1)^2(x-3)^2
(x^2+1)(x^2+9)
=x^4+9x^2+x^2+9
=x^4 - 4x^3 - 2x^2 + 12x + 9

More extensive working:
(x + 1)^2(x - 3)^2
= [(x + 1)(x - 3)]^2
= [x*x + 1*x - x*3 - 1*3]^2
= [x^2 + x - 3x - 3]^2
= [x^2 - 2x - 3]^2
= [x^2 - 2x - 3][x^2 - 2x - 3]
= x^2*x^2 - 2x*x^2 - 3*x^2 - x^2*2x + 2x*2x + 3*2x - x^2*3 + 2x*3 + 3*3
= x^4 - 2x^3 - 3x^2 - 2x^3 + 4x^2 + 6x - 3x^2 + 6x + 9
= x^4 - 2x^3 - 2x^3 - 3x^2 + 4x^2 - 3x^2 + 6x + 6x + 9
= x^4 - 4x^3 - 2x^2 + 12x + 9