A = \sqrt21(21 - 13)(21 - 14)(21 - 15) = \sqrt21 \cdot 8 \cdot 7 \cdot 6 - Treasure Valley Movers

Unlock the Power of Square Roots: Solve A = √[21(21 – 13)(21 – 14)(21 – 15)] with Confidence

Mathematics often presents challenges in the form of elegant algebraic expressions—like the powerful cancellation technique behind the square root equation:
A = √[21(21 – 13)(21 – 14)(21 – 15)]

At first glance, this may look like a standard expression, but it reveals a deeper mathematical insight involving factoring, simplification, and clever substitution. Let’s unpack this expression step-by-step to uncover how it simplifies effortlessly—and why it matters.

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Breaking Down the Expression: A = √[21(21 – 13)(21 – 14)(21 – 15)]

Start by calculating each term inside the parentheses:

• 21 – 13 = 8
  
• 21 – 14 = 7
  
• 21 – 15 = 6

So the expression becomes:
A = √[21 × 8 × 7 × 6]

Instead of leaving it as a product of four numbers, we simplify by reordering and grouping factors:

A = √(21 × 8 × 7 × 6) = √[21 × 7 × 8 × 6]

Notice the pairing: 21 and 7 are multiples, and 8 and 6 share common structure. This sets the stage for factoring to reveal perfect squares—the key to simplifying square roots.

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Step 1: Factor and Rearrange

Let’s factor each number into primes:

• 21 = 3 × 7
  
• 8 = 2³
  
• 7 = 7
  
• 6 = 2 × 3

Putting it all together:
A = √[(3 × 7) × (2³) × (7) × (2 × 3)]

Now combine like terms:

• 2³ × 2 = 2⁴ (since 2³ × 2 = 2⁴ = 16)
  
• 3 × 3 = 3²
  
• 7 × 7 = 7²

So the radicand becomes:
A = √(2⁴ × 3² × 7²)

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Step 2: Extract Perfect Squares

Using the property √(a×b) = √a × √b, we separate each squared term:

A = √(2⁴) × √(3²) × √(7²)