The Distributive Property: Free Response

The area of a rectangle is its height times its width, and the strip is $6$ units tall and $17$ units wide:

$6 \times 17 = 102.$

The strip covers $102$ square units.

Neither piece lost any height: the cut ran from the top edge straight to the bottom edge, so both pieces are still $6$ units tall. Only the widths differ, $12$ for the left piece and $5$ for the right.

$6 \times 12 = 72, \qquad 6 \times 5 = 30.$

Adding the two pieces accounts for every part of the card:

$72 + 30 = 102.$

Cutting is not the same as removing. The blade moves no card and leaves no gap, so the two pieces taken together are exactly the original strip. Whatever number of square units the strip covered, the pieces cover that same number, and this is settled before any multiplying begins.

Now read each route as a measurement of that one piece of card. Multiplying the height by the full width measures the strip in one go. Multiplying the height by $12$, then by $5$, and adding measures the same card in two goes, the left part and then the right. Both must report the same surface:

$6 \times 17 = 6 \times 12 + 6 \times 5.$

That is the distributive property, with the width $17$ written as $12 + 5$. Nothing in the reason used the numbers $12$ and $5$. A cut anywhere along the width splits it into two widths that add back to the full width, and the areas still have to total the same, so the rule holds for every split. The height is just as free: a taller strip only repeats the argument with a different first factor.

Adding inside the parentheses first:

$5 \times (4 + 3) = 5 \times 7 = 35.$

Splitting the $5$ across the sum instead:

$5 \times 4 + 5 \times 3 = 20 + 15 = 35.$

The two routes agree. That agreement is the distributive property.

Multiplying inside the parentheses first:

$5 \times (4 \times 3) = 5 \times 12 = 60.$

Splitting the $5$ into both factors instead:

$(5 \times 4) \times (5 \times 3) = 20 \times 15 = 300.$

These do not agree, and $300$ is exactly $5$ times $60$. Splitting is not allowed here.

Part A matches the rule exactly. A rectangle $5$ tall and $4 + 3$ wide can be cut across its width into a $5$ by $4$ piece and a $5$ by $3$ piece, and the two areas add back to the whole. That is why $5 \times 4 + 5 \times 3$ returns the same $35$.

Part B has no sum inside the parentheses. The rule needs two numbers that are added, and $4 \times 3$ is a single product, not two pieces adding to something, so the rule does not apply. Multiplying the $4$ by $5$ and the $3$ by $5$ puts the $5$ in twice:

$(5 \times 4) \times (5 \times 3) = 5 \times 5 \times 4 \times 3.$

That is $5 \times 60$, not $60$, which is exactly the factor of five between $300$ and the correct $60$.

Work one crate at a time first. A crate arrives holding $25$ notebooks and $4$ of them cannot be sold, so a single crate yields $25 - 4$ sellable notebooks. Every crate is identical, so there are $8$ such yields:

$8 \times (25 - 4).$

The parentheses carry the meaning here. They say that the subtraction is a fact about one crate, and that all $8$ crates deliver that same yield.

Taking the parentheses first, one crate yields $25 - 4 = 21$ sellable notebooks, and eight crates yield

$8 \times 21 = 168.$

Distributing instead multiplies both numbers inside by $8$, and the two products are joined by the same minus sign that sat inside the parentheses:

$8 \times 25 - 8 \times 4 = 200 - 32 = 168.$

Both routes report $168$ sellable notebooks.

Read each product back into the shop. The product $8 \times 25 = 200$ counts every notebook that came off the truck, sellable or not, as though nothing had been damaged. The product $8 \times 4 = 32$ counts the damaged notebooks, $4$ from each of the $8$ crates, pooled into one pile.

$200 - 32 = 168.$

The subtraction removes that pile all at once. The parentheses route removes the same notebooks $4$ at a time, one crate at a time, before any total is worked out. Both routes remove the same notebooks and keep the same notebooks, so they cannot report different counts.

Distributing the $8$ does not change the shop. It changes only the order of the counting, from crate by crate to all notebooks first and all damage second.

The round number just above $48$ is $50$, and the gap is $2$, so $48 = 50 - 2$. Sending the $7$ to both numbers gives one easy product and one small one:

$7 \times 48 = 7 \times 50 - 7 \times 2 = 350 - 14.$

Subtracting leaves

$350 - 14 = 336.$

Both products were easy to hold in mind: a multiple of $50$, and a single-digit product.

A convenient split of $63$ is $60 + 3$, since multiplying by $60$ and by $3$ are both easy. Distributing the $8$ across both parts,

$8 \times 63 = 8 \times 60 + 8 \times 3 = 480 + 24.$

Adding gives

$480 + 24 = 504.$

Another split works too, so long as its two parts recover $63$. That may be by adding, as here, or by subtracting: splitting as $70 - 7$ gives $560 - 56 = 504$, the same answer. Both are correct, and the first is easier because $480 + 24$ takes less work than $560 - 56$.

Follow the chain one expression at a time. Replacing $48$ by $50 - 2$ is sound, because $50 - 2$ really is $48$, so $7 \times (50 - 2)$ is the same product, just written differently.

The next expression is where it breaks. The outside factor has to reach every number inside the parentheses, and here only the $50$ was multiplied; the $2$ was copied down untouched. The rule gives two products, and the second one is $7 \times 2 = 14$, not $2$:

$7 \times 48 = 7 \times 50 - 7 \times 2 = 350 - 14 = 336.$

The size of the error is itself a check. Subtracting $2$ instead of $14$ takes away $12$ too little, so the student answer is $12$ greater than the correct one: $348 - 336 = 12$.

Both products are built on the same factor $7$, so the rule read from right to left takes that $7$ out in front and leaves the two other numbers inside one pair of parentheses:

$7 \times 13 + 7 \times 7 = 7 \times (13 + 7).$

The inside now adds to a round number:

$7 \times 20 = 140.$

The collected form is less work: one multiplication by a round number instead of two awkward products and an addition. Multiplying out the long way gives $91 + 49 = 140$ as well, so that route is a check rather than a shortcut.

Look for the largest number that divides both terms. The numbers dividing $45$ are $1, 3, 5, 9, 15, 45$, and those dividing $72$ are $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$. The largest they share is $9$, and $45 = 9 \times 5$ while $72 = 9 \times 8$, so the shared $9$ comes out in front:

$45 + 72 = 9 \times 5 + 9 \times 8 = 9 \times (5 + 8).$

Multiplying back out is the check: $9 \times 5 = 45$ and $9 \times 8 = 72$, which is the sum we began with. The factoring is finished, because $5$ and $8$ share no factor above $1$, so there is nothing left inside to pull out. If you want the total as well, either side comes to $117$.

There are two separate checks here. Distribute back to test whether the equation is true. Then look at the two numbers inside the parentheses to test whether the factoring is finished. An equation can pass the first check and fail the second.

Is the equation true? Multiply the right side back out. $2 \times 15 = 30$ and $2 \times 21 = 42$, so the right side is $30 + 42$, exactly the left side. Both sides come to $72$. There is nothing wrong with the equation.

Is the factoring finished? Look at what the two numbers inside the parentheses still share. Both $15$ and $21$ are divisible by $3$, so a common factor is still sitting inside. Taking that out as well collects a $6$ in front:

$30 + 42 = 6 \times (5 + 7).$

Now $5$ and $7$ share no factor above $1$, so nothing remains inside and the factoring is complete. Checking, $6 \times 5 = 30$ and $6 \times 7 = 42$. Notice that the classmate's $2$ and the extra $3$ multiply to that $6$, which is why the two answers agree rather than contradict each other. The classmate started correctly and stopped one step early.