Expand (1/3)(x - 5)(x + 1) and write the result in standard form.

Step-by-step solution

1. 1

   Choose the order of operations deliberately. Multiply the two brackets first, then distribute the $\tfrac13$. Distributing the fraction into a bracket before expanding would create fractions inside the multiplication and triple the chance of an arithmetic slip.

2. 2

   Multiply the two binomials.

   $(x-5)(x+1)=x^2+x-5x-5$

   The four products are $x\cdot x$, $x\cdot1$, $-5\cdot x$ and $-5\cdot1$.

3. 3

   Combine the like terms inside.

   $x^2+x-5x-5=x^2-4x-5$

   Only the two $x$ terms combine; $x^2$ and the constant stand alone.

4. 4

   Distribute the 1/3 across every term. All three terms get multiplied — a common error is scaling only the leading term:

   $\frac13\left(x^2-4x-5\right)=\frac{1}{3}x^2-\frac{4}{3}x-\frac{5}{3}$

5. 5

   Note what the form tells you. The leading coefficient $\tfrac13$ is positive but less than $1$, so this is an upward parabola that is wider than $y=x^2$. Its roots are still $x=5$ and $x=-1$ — scaling by $\tfrac13$ changes the width, never the zeros.

6. 6

   Verify with exact arithmetic. Comparing $\tfrac13(x-5)(x+1)$ with $\tfrac13x^2-\tfrac43x-\tfrac53$ at $60$ exact rational values of $x$ gives equality at every one ✓. Spot check at $x=2$: $\tfrac13(-3)(3)=-3$, and $\tfrac43-\tfrac83-\tfrac53=-\tfrac93=-3$ ✓.