Matematica — Liceo

The method of shapes

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tagsmatematica, classe/prima, topic/calcolo-letterale, concept/prodotti-notevoli, concept/quadrato-di-binomio, skill/usare-formule, skill/calcolare

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Observation — Method of shapes

To apply the special-product formulas, geometric shapes are assigned to the terms: the first term is the “circle” ◯\textcolor{#1f77b4}{\bigcirc}, the second is the “square” □\textcolor{#d62728}{\square}.

Example: (2−3ab)3=(◯+□)3(2-3ab)^3 = (\textcolor{#1f77b4}{\bigcirc}+\textcolor{#d62728}{\square})^3 with ◯=2\textcolor{#1f77b4}{\bigcirc}=2 and □=−3ab\textcolor{#d62728}{\square}=-3ab. =◯3+3 ◯2 □+3 ◯ □2+□3= \textcolor{#1f77b4}{\bigcirc}^3 + 3\,\textcolor{#1f77b4}{\bigcirc}^2\,\textcolor{#d62728}{\square} + 3\,\textcolor{#1f77b4}{\bigcirc}\,\textcolor{#d62728}{\square}^2 + \textcolor{#d62728}{\square}^3 Substituting: =23+3⋅4⋅(−3ab)+3⋅2⋅9a2b2+(−3ab)3=8−36ab+54a2b2−27a3b3= 2^3 + 3\cdot 4\cdot(-3ab) + 3\cdot 2\cdot 9a^2b^2 + (-3ab)^3 = 8 - 36ab + 54a^2b^2 - 27a^3b^3 The same method works for (A+B)2(A+B)^2, (A+B)(A−B)(A+B)(A-B) and (A+B+C)2(A+B+C)^2: you draw the shapes in place of the terms, then substitute.

The method of shapes in reverse (recognition). Given a trinomial, I can check whether it is a square of a binomial using the shapes in reverse:

  1. Is the first term ◯2\textcolor{#1f77b4}{\bigcirc}^2? ⇒\Rightarrow I find ◯=primo termine\textcolor{#1f77b4}{\bigcirc} = \sqrt{\text{primo termine}}.
  2. Is the last term □2\textcolor{#d62728}{\square}^2? ⇒\Rightarrow I find □=ultimo termine\textcolor{#d62728}{\square} = \sqrt{\text{ultimo termine}}.
  3. Is the middle term =2⋅◯⋅□= 2\cdot\textcolor{#1f77b4}{\bigcirc}\cdot\textcolor{#d62728}{\square}? If so, the trinomial is (◯+□)2(\textcolor{#1f77b4}{\bigcirc}+\textcolor{#d62728}{\square})^2. ✓

Example. 16a2b2+16abc+4c216a^2b^2+16abc+4c^2. I write (◯+□)2=◯2+2◯□+□2(\textcolor{#1f77b4}{\bigcirc}+\textcolor{#d62728}{\square})^2 = \textcolor{#1f77b4}{\bigcirc}^2 + 2\textcolor{#1f77b4}{\bigcirc}\textcolor{#d62728}{\square}+\textcolor{#d62728}{\square}^2.

  • ◯2=16a2b2⇒◯=4ab\textcolor{#1f77b4}{\bigcirc}^2 = 16a^2b^2 \Rightarrow \textcolor{#1f77b4}{\bigcirc}=4ab.
  • □2=4c2⇒□=2c\textcolor{#d62728}{\square}^2 = 4c^2 \Rightarrow \textcolor{#d62728}{\square}=2c.
  • Check: 2⋅◯⋅□=2⋅4ab⋅2c=16abc2\cdot\textcolor{#1f77b4}{\bigcirc}\cdot\textcolor{#d62728}{\square} = 2\cdot 4ab\cdot 2c = 16abc. ✓

Therefore 16a2b2+16abc+4c2=(4ab+2c)216a^2b^2+16abc+4c^2 = \boxed{(4ab+2c)^2}.

Topics: Algebraic calculus
Concepts: Special products · Square of a binomial
Skills: Calculating · Using formulas

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