Use the binomial theorem to expand the binomial.

Expand the following:
(x/2+t)^4

 

(t+x/2)^4 = sum_(k=0)^4 binomial(4, k) (x/2)^(4-k) t^k = binomial(4, 0) (x/2)^4 t^0+binomial(4, 1) (x/2)^3 t^1+binomial(4, 2) (x/2)^2 t^2+binomial(4, 3) (x/2)^1 t^3+binomial(4, 4) (x/2)^0 t^4:
(binomial(4, 0) x^4)/16+1/8 binomial(4, 1) t x^3+1/4 binomial(4, 2) t^2 x^2+1/2 binomial(4, 3) t^3 x+binomial(4, 4) t^4

 

binomial(4, 0) = 1, binomial(4, 1) = 4, binomial(4, 2) = 6, binomial(4, 3) = 4 and binomial(4, 4) = 1:
t^4+(4 t^3 x)/2+6 t^2 (x/2)^2+4 t (x/2)^3+(x/2)^4

 

4/2 = (2×2)/2 = 2:
t^4+2 t^3 x+6 t^2 (x/2)^2+4 t (x/2)^3+(x/2)^4

 

Multiply each exponent in x/2 by 2:
t^4+2 t^3 x+6 t^2 (1/2)^2 x^2+4 t (x/2)^3+(x/2)^4

 

(1/2)^2 = 1^2/2^2:
t^4+2 t^3 x+1^2/2^2 6 t^2 x^2+4 t (x/2)^3+(x/2)^4

 

1^2 = 1:
t^4+2 t^3 x+(6 t^2 x^2)/2^2+4 t (x/2)^3+(x/2)^4

 

2^2 = 4:
t^4+2 t^3 x+(6 t^2 x^2)/4+4 t (x/2)^3+(x/2)^4

 

The gcd of 6 and 4 is 2, so (t^2 x^2×6)/4 = ((2×3) t^2 x^2)/(2×2) = 2/2×(3 t^2 x^2)/2 = (3 t^2 x^2)/2:
t^4+2 t^3 x+(3 t^2 x^2)/2+4 t (x/2)^3+(x/2)^4

 

Multiply each exponent in x/2 by 3:
t^4+2 t^3 x+(3 t^2 x^2)/2+4 t (1/2)^3 x^3+(x/2)^4

 

(1/2)^3 = 1^3/2^3:
t^4+2 t^3 x+(3 t^2 x^2)/2+1^3/2^3 4 t x^3+(x/2)^4

 

1^3 = 1:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/2^3+(x/2)^4

 

2^3 = 2×2^2:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/2×2^2+(x/2)^4

 

2^2 = 4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/(2×4)+(x/2)^4

 

2×4 = 8:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/8+(x/2)^4

 

4/8 = 4/(4×2) = 1/2:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+(x/2)^4

 

Multiply each exponent in x/2 by 4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+(1/2)^4 x^4

 

(1/2)^4 = 1^4/2^4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+1^4/2^4 x^4

 

1^4 = 1:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+x^4/2^4

 

2^4 = (2^2)^2:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+x^4/(2^2)^2

 

2^2 = 4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+x^4/4^2

 

4^2 = 16:
Answer: |t^4 + 2t^3 x + (3t^2x^2)/2 +( tx^3)/2 + x^4/16