Divide (4a² − b²)/(b + 2a) by (b − 2a)² and simplify.

Step-by-step solution

1. 1

   Factor the difference of squares and write the divisor as a fraction.

   $4a^{2}-b^{2}=(2a-b)(2a+b),\qquad (b-2a)^{2}=\frac{(b-2a)^{2}}{1}$

   so the expression becomes

   $\frac{(2a-b)(2a+b)}{b+2a}\cdot\frac{1}{(b-2a)^{2}}$

2. 2

   Cancel the sum factor. Addition is commutative, so $b+2a=2a+b$ and the two cancel outright:

   $\frac{2a-b}{(b-2a)^{2}}$

3. 3

   Use the fact that a square is unchanged by a sign flip. Since $(b-2a)^{2}=(2a-b)^{2}$, the expression is

   $\frac{2a-b}{(2a-b)^{2}}$

   This rewriting is the key step: it turns the sign mismatch into an exact cancellation.

4. 4

   Cancel one factor of (2a − b).

   $\frac{1}{2a-b}$

   $\boxed{\dfrac{1}{2a-b}}\qquad(2a\neq\pm b)$

5. 5

   Check with a test value. At $a=1$, $b=0$: the original is $\dfrac{4}{2}\div 4=2\div 4=\tfrac12$, and the answer gives $\tfrac{1}{2}$ ✓. At $a=1$, $b=1$: original $=\dfrac{3}{3}\div 1=1$, answer $=\tfrac{1}{1}=1$ ✓.

6. 6

   Note the equivalent form. The answer can equally be written $-\dfrac{1}{b-2a}$; the two differ only in how the sign is carried, so either is acceptable — but writing $\dfrac{1}{b-2a}$ without the minus would be wrong.