Multi-Digit Multiplication: The Standard Algorithm

$128$

$30 \times 4 = 120$ and $2 \times 4 = 8$, so $128$.

$560$

One factor of ten shifts every digit one place left.

$276$

$23 \times 2 = 46$ and $23 \times 10 = 230$, so $46 + 230 = 276$.

$900$

$45 \times 2 = 90$, then shift once for the ten: $900$.

About $800$

$40 \times 20 = 800$. The exact answer is $819$.

$1600$

$64 \times 5 = 320$ and $64 \times 20 = 1280$, so $320 + 1280 = 1600$.

$1260$

$105 \times 2 = 210$ and $105 \times 10 = 1050$, so $210 + 1050 = 1260$.

One row per digit.

Three

The units, tens and hundreds digits each give a partial product, shifted zero, one and two places respectively.

$6848$

$214 \times 2 = 428$ and $214 \times 30 = 6420$, so $428 + 6420 = 6848$.

Estimate: $200 \times 30 = 6000$, and $6848$ is in that neighborhood.

Estimate the answer first.

She did not shift the second row. The answer is $1904$.

Both partial products are correct as bare multiplications: $56 \times 4 = 224$ and $56 \times 3 = 168$.

But the $3$ in $34$ is three tens, so the second row is $56 \times 30 = 1680$, not $168$.

$224 + 1680 = 1904$.

An estimate catches it immediately: $60 \times 30 = 1800$, and her $392$ is not remotely near that. Being out by roughly a factor of five is the signature of one row sitting a place too far right.