Multiplying Two-Digit Numbers × Two-Digit Numbers (Long Multiplication)

Introduction & Key Vocabulary

• Video Source: “Math with Mr. J.”
• Objective: Demonstrate how to multiply any two-digit number by another two-digit number using the traditional (long-multiplication) algorithm.
• Key Terms
  • Product – the answer to a multiplication problem.
  • Partial Product – the intermediate result obtained when one digit of the multiplier is used to multiply the entire multiplicand; all partial products are added to obtain the final product.
  • Place Value Alignment – lining up ones under ones, tens under tens, etc., to prevent place-value errors.
  • Carrying (Regrouping) – moving values greater than 9 to the next higher place value.
  • Placeholder Zero – a $0$ inserted in the second (and subsequent) row(s) of multiplication to show that the digit being used is in the tens (or hundreds, thousands, …) place.

General Procedure (Long Multiplication)

1. Write the Multiplication Vertically
   • Ensure digits in corresponding place values are directly above/below one another.
2. Multiply the Ones-Place Digit of the Bottom Number by Every Digit of the Top Number
   • Record each result; when the product is ≥10, write the ones digit and carry the tens digit.
3. Cross Off Completed Digits & Carries (Optional but recommended)
   • Prevents accidental reuse.
4. Move to the Tens-Place Digit of the Bottom Number
   • Insert one pointer placeholder zero beneath the ones column.
   • Multiply and carry as before.
5. Repeat for Additional Digits (Hundreds, Thousands, …)
   • Each new row receives another placeholder zero.
6. Add All Partial Products to obtain the final product.

Mathematically, for $\text{AB}\times\text{CD}$ (where $\text{A,B,C,D}$ are digits, $A,C\neq0$):

\text{AB}\times\text{CD} = (10A+B)(10C+D) = (10A+B)D + (10A+B)\,(10C) \
= D(10A+B) + 10C(10A+B)

The first term is the “ones-row” partial product; the second term requires a placeholder zero and is the “tens-row” partial product.

Example 1 – $39 \times 24$

• Setup

     39
  ×  24
  -----

• Ones are aligned ($9$ over $4$); tens are aligned ($3$ over $2$).
  • Step 1: Multiply by the Ones-Place (4)
• $4\times9 = 36$ → write $6$, carry $3$.
• $4\times3 = 12$ (but $3$ really means $30$; value-wise we are doing $4\times30=120$). Add carried $3$: $12+3=15$. Write $15$ (
  the $1$ is in the hundreds place).
• First partial product: 156.
  • Step 2: Placeholder Zero
• Because the next digit $2$ is really $20$, insert a $0$ under the ones column.
  • Step 3: Multiply by the Tens-Place (2)
• $2\times9 = 18$ → write $8$ in tens column, carry $1$ to hundreds.
• $2\times3 = 6$; add carry $1$ → $7$.
• Second partial product: 780.
  • Step 4: Add Partial Products
    
    \begin{array}{r}
    156\
    +780\
    -----\
    936
    \end{array}
    
• Column-wise addition:
  • Ones: $6+0=6$.
  • Tens: $5+8=13$ → write $3$, carry $1$.
  • Hundreds: $1+1=2$; $2+7=9$ (after adding carry).
  • Final Product: $39\times24 = 936$.

Example 2 – $68 \times 57$

• Setup

     68
  ×  57
  -----

• Step 1: Multiply by the Ones-Place (7)
  1. $7\times8 = 56$ → write $6$, carry $5$.
  2. $7\times6 = 42$; add carry $5$ ⇒ $47$.
  • First partial product: 476.
• Step 2: Placeholder Zero
  • The next digit $5$ is in the tens place (value $50$); write one $0$ below the ones column before continuing.
• Step 3: Multiply by the Tens-Place (5)
  1. $5\times8 = 40$ → write $0$ (under tens), carry $4$.
  2. $5\times6 = 30$; add carry $4$ ⇒ $34$.
  • Second partial product: 3400.
• Step 4: Add Partial Products$\begin{array}{r} ~~~476\ +3400\ -----\ 3876 \end{array}$
  • Ones: $6+0=6$.
  • Tens: $7+0=7$.
  • Hundreds: $4+4=8$.
  • Thousands: $3$ (no addition needed).
• Final Product: $68\times57 = 3\,876$.

Strategy, Tips & Common Errors

• Align Digits Carefully: Even a single-column misalignment can invalidate the entire computation.
• Carry Immediately: Write the carry digit above the next column as soon as it is produced.
• Cross Off Used Carries & Digits: Keeps the workspace clear—especially important on multi-step problems.
• Insert Exactly One Placeholder Zero per Place-Value Shift: Two-digit multiplier ⇒ one zero; three-digit multiplier ⇒ up to two zeros, etc.
• Double-Check Each Partial Product: A mistake in an early step propagates into the final sum.

Connections & Significance

• Foundation for Larger Multiplications: The exact same framework extends to three-digit × three-digit numbers (add more rows and placeholder zeros).
• Link to the Distributive Property
  • Example 1 illustrated $39\times24=(30+9)(20+4)$.
  • Expanded form: $39\times24=30\times20 + 30\times4 + 9\times20 + 9\times4$, though the long-multiplication algorithm groups these terms into two partial products for efficiency.
• Real-World Relevance: Mental or paper-based long multiplication is valuable when calculators are unavailable (e.g.
  standardized tests, quick estimations, or demonstrating arithmetic fluency).
• Pedagogical Note: Visual aids (grid or area models) often precede this algorithm to build conceptual understanding of place value and the distributive property.

Summary

• Long multiplication requires sequential multiplication of each digit in the multiplier with the entire multiplicand.
• Place-value awareness (placeholder zeros, correct alignment) is crucial.
• The algorithm breaks a seemingly complex multiplication into manageable, repeatable steps, culminating in an addition of partial products.
• Mastery of this method directly supports higher-level arithmetic and algebraic manipulation.