Expand (x - 2/root x)^4.

Step-by-step solution

1. 1

   Set up the binomial pattern. For a fourth power the coefficients are the fifth row of Pascal's triangle, $1,4,6,4,1$, with alternating signs because the second term is subtracted:

   $(A-B)^4=A^4-4A^3B+6A^2B^2-4AB^3+B^4$

   Take $A=x$ and $B=\dfrac{2}{\sqrt{x}}=2x^{-1/2}$. Rewriting $B$ with a fractional exponent up front is what keeps the arithmetic straightforward.

2. 2

   Note how the exponents will step. Each term trades one factor of $A$ (exponent $+1$) for one factor of $B$ (exponent $-\tfrac12$), so the exponent drops by $\tfrac32$ from term to term: $4,\ \tfrac52,\ 1,\ -\tfrac12,\ -2$. Knowing this in advance catches any slip.

3. 3

   Compute the first three terms.

   $A^4=x^4$

   $-4A^3B=-4x^3\cdot2x^{-1/2}=-8x^{5/2}$

   $6A^2B^2=6x^2\cdot\frac{4}{x}=24x$

   The third term is where the square root disappears entirely, since $B^2=4x^{-1}$.

4. 4

   Compute the last two terms.

   $-4AB^3=-4x\cdot\frac{8}{x^{3/2}}=-32x^{-1/2}=-\frac{32}{\sqrt{x}}$

   $B^4=\left(2x^{-1/2}\right)^4=\frac{16}{x^2}$

5. 5

   Assemble the expansion.

   $\left(x-\frac{2}{\sqrt{x}}\right)^4=x^4-8x^{5/2}+24x-\frac{32}{\sqrt{x}}+\frac{16}{x^2}$

   The domain is $x>0$, since $\sqrt{x}$ sits in a denominator.

6. 6

   Verify numerically at several values of x. At $x=1$ both the original and the expansion give $1$; at $x=0.3$ both give $126.1678$; at $x=2.7$ both give $4.83477$; at $x=5.5$ both give $466.40627$ ✓. Agreement at four well-spread points, including one below $1$, confirms every coefficient and exponent.