4 [(4x^2 + 4x)]^2 + 4(4x^2 + 4x)
I got up to:
16 ( 4x^4 + 8x^3 + 5x^2 + x)
Is this right and also the domain is X is all real numbers? (-infinity, infinity)?
I could probably factor another x out:
16x ( 4x^3 + 8x^2 + 5x + 1)
So X | E R?
To simplify the expression 4 [(4x^2 + 4x)]^2 + 4(4x^2 + 4x), you can follow these steps:
Step 1: Apply the distributive property inside each square bracket:
4 [16x^4 + 32x^3 + 16x^2] + 4(4x^2 + 4x)
Step 2: Multiply each term inside the square bracket by 4:
64x^4 + 128x^3 + 64x^2 + 16x^2 + 16x
Step 3: Combine like terms:
64x^4 + 128x^3 + 80x^2 + 16x
So, the simplified expression is: 64x^4 + 128x^3 + 80x^2 + 16x
As for the domain, x can indeed take any real value since there are no restrictions or limitations given in the original expression. Therefore, the domain is indeed (-∞, ∞), meaning that x can be any real number.