Factor the expression x^2/9 minus the square of (2x/3 - m)/4.

Step-by-step solution

1. 1

   Recognise both terms as perfect squares. The first is $\dfrac{x^2}{9}=\left(\dfrac{x}{3}\right)^2$ because $9=3^2$; the second is written as a square already. So with

   $a=\frac{x}{3},\qquad b=\frac{\frac23x-m}{4}$

   the expression is exactly $a^2-b^2$. Expanding first would work but creates fractions with denominators $9$ and $16$ for no reason.

2. 2

   Simplify $b$ before applying the identity. Splitting the numerator,

   $b=\frac{\frac23x}{4}-\frac{m}{4}=\frac{x}{6}-\frac{m}{4}$

   Dividing $\tfrac23x$ by $4$ gives $\tfrac{x}{6}$, not $\tfrac{x}{12}$ — a frequent arithmetic slip.

3. 3

   Apply $a^2-b^2=(a-b)(a+b)$ and simplify each factor.

   $a-b=\frac{x}{3}-\frac{x}{6}+\frac{m}{4}=\frac{x}{6}+\frac{m}{4}=\frac{2x+3m}{12}$

   $a+b=\frac{x}{3}+\frac{x}{6}-\frac{m}{4}=\frac{x}{2}-\frac{m}{4}=\frac{2x-m}{4}$

4. 4

   Multiply the two factors.

   $\frac{2x+3m}{12}\cdot\frac{2x-m}{4}=\frac{(2x+3m)(2x-m)}{48}$

5. 5

   Check numerically. At $x=3$, $m=1$: the original is $\tfrac99-\left(\tfrac{2-1}{4}\right)^2=1-\tfrac1{16}=0.9375$; the factored form gives $\tfrac{(6+3)(6-1)}{48}=\tfrac{45}{48}=0.9375$ ✓. At $x=-2$, $m=5$: original $=\tfrac49-\left(\tfrac{-4/3-5}{4}\right)^2=0.4444-2.5069=-2.0625$; factored $=\tfrac{(-4+15)(-4-5)}{48}=\tfrac{-99}{48}=-2.0625$ ✓.