01.046 – Matemática – Álgebra, Produtos notáveis. Exercícios resolvidos.

Exercícios de produtos notáveis.

  1. Usando a regra do quadrado da soma de dois números, obtenha os trinômios quadrados perfeitos que resultam das expressões a seguir. a)$\color{Orchid}{{(uv + z)}^2}$;  b)$\color{Orchid}{{(5m + r)}^2}$;  c)$\color{Orchid}{{(7 + 2p)}^2}$; d)$\color{Orchid}{{(a + 6b)}^2}$; e)$\color{Orchid}{{(10x^{2 }+ y^{2})}^2}$; f)$\color{Orchid}{{(mp^{3} + nr^{2})}^2}$.

a)\begin{align}{(uv + z)}^2 \\ = {(uv)^2 + 2\cdot{uv}\cdot{z} + {z}^2} & =  {u^2 v^2} + 2uvz + z^2\end{align}

b)\begin{align}{(5m + r)}^2 \\ =  (5m)^{2} + 2\cdot{5m}\cdot r + r^{2} & =  25m^2 +10mr + r^2\end{align}

c)\begin{align} {(7 + 2p)}^2 \\ = 7^2 + 2\cdot 7\cdot {2p} + {^(2p)}^2 & = 49 + 28p +4p^2\end{align}

d)\begin{align}{(a + 6b)}^2 \\= { a^2 + 2\cdot a\cdot{6b} + {(6b)}^2 } & =  a^2 + 12ab + 36b^2\end{align}

e)\begin{align}{(10x^{2 }+ y^{2})}^2 \\  =  {(10x^{2})}^2 + 2\cdot{(10x^{2}}\cdot{y^{2}} + {y^{2}}^2 & =   100x^4 + 20x^{2}y^{2} + y^{4}\end{align}

f)\begin{align}{(mp^{3} + nr^{2})}^2 \\= {(mp^{3})}^2 +2\cdot{(mp^3)}\cdot (nr^{2}) + {(nr^{2})}^2 & =  m^{2}p^{6} + 2mnp^{3}r^{2} + n^{2}r^{4}\end{align}

2. Faça o mesmo usando a regra do quadrado da diferença entre dois números, com as expressões abaixo. a)$\color{Sepia}{{(5a – 2b)}^2}$; b)$\color{Sepia}{{(a^{2}i – b^{3}j)}^2}$; c)$\color{Sepia}{{(2vx – 3uy)}^2}$; d)$\color{Sepia}{{(4 q^{3} – 6p^{2})}^{2}}$; e)$\color{Sepia}{{(12 – 3 a^{3})}^2}$; f)$\color{Sepia}{{(15 – 3x)}^2}$; g)$\color{Sepia}{{(7x – 8y)}^2}$

a)\begin{align}{(5a – 2b)}^2 \\ =  {(5a)}^2 – 2\cdot {5a}\cdot{2b} +{(ab)}^2 & =  25a^2 – 20ab + 4b^2\end{align}

b)\begin{align}{(a^{2}i – b^{3}j)}^2 \\  =  [(a^2)i]^2 – 2\cdot{a^2}i\cdot{b^3} + {(b^3)}^2 & =  a^{4}i^{2} – 2a^{2}b^{3}i + b^6 \end{align}

c)\begin{align}{(2vx – 3uy)}^2 \\ = {(2vx)^2 – 2\cdot {(2vx)}\cdot{(3uy)} + {(3uy)}^2} &  =  4v^{2}x^{2} – 12uvxy + 9u^{2}y^{2} \end{align}

d)\begin{align}{(4 q^{3} – 6p^{2})}^{2} \\ =  (4q^{3})^2 – 2\cdot (4q^{3})\cdot(6p^{2}) +{(6p^{2})}^2 \\ =  16q^6 – 48q^{3}p^{2} + 36p^{4}\end{align}

e)\begin{align}{(12 – 3 a^{3})}^2 \\ = {(12)^2 – 2\cdot{12}\cdot{3a^{3}} + {(3a^{3})}^2} & =  144 – 72a^{3} + 9a^6\end{align}

f)\begin{align}{(15 – 3x)}^2 \\  =  {(15)}^2 – 2\cdot {15}\cdot{(3x)} + {(3x)}^2 \\ =  225 – 90x + 9x^2 \end{align}

g)\begin{align}{(7x – 8y)}^2 \\ = {(7x)}^2 – 2\cdot{(7x)}\cdot 8y + {(8y)}^2 & = 49x^2 – 112xy + 64y^2\end{align}

3. Usando a regra do produto da soma de dois números pela sua diferença, obtenha os binômios resultantes das multiplicações abaixo. a)$\color{Indigo}{{(7 + 2x)}{(7 – 2x)}}$; b)$\color{Indigo}{{(5 – 3y)}{(5 + 3y)}}$; c)$\color{Indigo}{{(ab^{2} + b)}{(ab^{2} – b)}}$; d)$\color{Indigo}{{(xy + xz)}{(xy – xz)}}$; e)$\color{Indigo}{{(4m – 3n)}{(4m + 3n)}}$; f)$\color{Indigo}{{(7x^{3} + 2y^{2})}{(7x^{3} – 2y^{2})}}$.

a)\begin{align}{(7 + 2x)}{(7 – 2x)} \\ =  {7^2 – {(2x)}^2} &  = 49 – 4x^2\end{align}

b)\begin{align}{(5 – 3y)}{(5 + 3y)} \\ =  5^2 – {(3y)}^2 & =  25 – 9y^2\end{align}

c)\begin{align}{(ab^{2} + b)}{(ab^{2} – b)}\\ =  {(ab^{2}}^{2} – b^2&  =  a^{2}b^{4} – b{2}\end{align}

d)\begin{align}{(xy + xz)}{(xy – xz)} \\ =  {(xy)}^{2} – {(xz)}^{2} & =  x^{2}y^{2} – x^{2}z^{2}\end{align}

e)\begin{align}{(4m – 3n)}{(4m + 3n)}\\ =  {(4m)}^{2} – {(3n)}^{2} & =  16m^2 – 9n^2 \end{align}

f)\begin{align}{(7x^{3} + 2y^{2})}{(7x^{3} – 2y^{2})}\\ = {(7x^{3})}^{2} – {(2y^{2})}^2 & =   49x^{6} – 4y^{4}\end{align}

4. Use agora a regra do cubo da soma de dois números para obter os polinômios de quatro termos resultantes das expressões abaixo.  a)$\color{Indigo}{{(7 +2j)}^3}$; b)$\color{Indigo}{(x + 3yz)}^3;$ c)$\color{Indigo}{{(4l + 5m)}^3}$; d)$\color{Indigo}{{(ma + nb)}^3}$; e)$\color{Indigo}{{(11 + 4r)}^3}$.

a)\begin{align}{(7 +2j)}^3\\  = 7^3 + 3\cdot {7^{2}}\cdot{2j} + 3\cdot {7}\cdot {(2j)}^{2} + {(2j)}^3 & = 343 + 294j + 84j^2  + 8j^3\end{align}

b)\begin{align}{(x + 3yz)}^3\\  = x^3 + 3\cdot x^{2}\cdot {(3yz)} + 3\cdot x\cdot {(3yz)}^2 + {(3yz)}^3&  =  x^3 + 9x^{2}yz + 27xy^{2}z^{2} + 27y^{3}z^{3}\end{align}

c)\begin{align}{(4l + 5m)}^3\\ = {(4l)}^3 +3\cdot{4l}^{2}\cdot{(5m)} +3\cdot{(4l)}\cdot{(5m)}^2 + {(5m)}^3 & =  64l^3 + 240l^{2}m + 125m^{3}\end{align}

d)\begin{align}{(ma + nb)}^3 \\ =  {(ma)^3} + 3\cdot {(ma)}^{2}\cdot {(nb)} +3\cdot {(ma)}\cdot {(nb)}^{2} + {(nb)}^{3} & =  m^{3}a^{3} + 3m^{2}na^{2}b + 3mn^{2}ab^{2} + n^{3}b^{3} \end{align}

e)\begin{align}{(11 + 4r)}^3 \\ = 11^3 + 3\cdot 11^{2}\cdot{(4r)} + 3\cdot 11\cdot{(4r)}^{2} + {(4r)}^{3} & =  1331 + 1452 r + 528 r^2 + 64 r^3\end{align}

5. Vamos fazer o mesmo com a regra do cubo da diferença. a)$\color{Orchid}{{(4m – 2)}^3}$; b)$\color{Orchid}{{(3x – 5y)}^3}$; c)$\color{Orchid}{{(9 – 5a)}^3}$; d)$\color{Orchid}{{(5 – 4x)}^3}$; e)$\color{Orchid}{{(10 – 5c)}^3}$; f)$\color{Orchid}{{(3ab – x)}^3}$; g$\color{Orchid}{{(pq^{2} – rq)}^3}$.

a)\begin{align}{(4m – 2)}^3 \\ =  {(4m)}^{3} – 3\cdot {(4m)^{2}}\cdot 2 + 3\cdot{(4m)}\cdot 2^{2} – 2^{3} & =  64m^{3} – 96m^{2} + 48m – 8\end{align}

b)\begin{align}{(3x – 5y)}^3 \\ = (3x)^{3} – 3\cdot (3x)^{2}\cdot {(5y)} + 3\cdot{(3x)}\cdot (5y)^{2} – {(5y)}^{3} & = 27x^3 – 135x^{2}y + 225xy^{2} – 125y^{3}\end{align}

c)\begin{align}{(9 – 5a)}^3 \\ =  {(9)^{3}} – 3\cdot (9)^2\cdot (5a) + 3\cdot 9 \cdot(5a)^{2} -(5a)^{3} & = 729 – 1215 a + 675 a^2 – 125 a^3\end{align}

d)\begin{align}{(5 – 4x)}^3 \\ =  5^{3} – 3\cdot 5^{2}\cdot (4x) + 3\cdot 5\cdot (4x)^{2} – {(4x)}^{3} & =   125 – 300x + 120 x^{2} – 64x^{3}\end{align}

e)\begin{align}{(10 – 5c)}^3 \\ = 10^{3} – 3\cdot(5)^{2}\cdot (5c) + 3\cdot{10}\cdot {(5c)}^{2} – {(5c)}^{3} & =  1000 – 375 c + 750 c^{2} – 125c^{3}\end{align}

f)\begin{align}{(3ab – x)}^3 \\ = {(3ab)}^{3} – 3\cdot {(3ab)}^{2}\cdot x + 3\cdot{(3ab)}\cdot x^{2} – x^{3} & =  27a^{3}b^{3} – 27a^{2}b^{2}x + 9 abx^{2} – x^{3} \end{align}

g)\begin{align}{(pq^{2} – rq)}^3 \\=  {(pq^{2})}^{3} – 3\cdot {(pq^{2})}^{2}\cdot rq + 3\cdot {(pq^{2})}\cdot {(rq)}^{2} – {(rq)}^{3} & =   p^{3}q^{6} – 3p^{2}q^{5}r + 3pq^{4}r^{2} – q^{3}r^{3}\end{align}

6. Chegou o momento de usar as regras mais avançadas. Multiplique os quadrados das somas pelas diferenças dos mesmos números, usando a regra vista no post anterior. a)$\color{Orchid}{{(ax + by)}^{2}\cdot {(ax – by)}}$; b)$\color{Orchid}{{(5 + 3x)}^{2}\cdot{(5 – 3x)}}$; c)$\color{Orchid}{{(4n + m^{2})}^{2}\cdot{(4n – m)}}$; d)$\color{Orchid}{{(5a + 3b)}^{2}\cdot{(5a – 3b)}} $; e)$\color{Orchid}{{(7x + 2y)}^{2}\cdot{7x -2y)}}$; f)$\color{Orchid}{{(10 + 3v)}^{2}\cdot{(10 – 3v)}}$; g)$\color{Orchid}{{(px + qy)}^{2}\cdot{(px – qy)}}$.

a)\begin{align}{(ax + by)}^{2}\cdot {(ax – by)}\\  = {(ax)}^3 + {(ax)}^{2}\cdot{(by)} – ax\cdot {(by)}^{2} – {(by)}^{3}&  = a^{3}x^{3} + a^{2}bx^{2}y – ab^{2}xy^{2} – {(by)}^{3}\\  = a^{3}x^{3} + a^2bx^2y – ab^2xy^2 – b^3y^3\end{align}

b)\begin{align}{(5 + 3x)}^{2}\cdot{(5 – 3x)} \\ = 5^{3} + 5^2\cdot{3x} – 5\cdot {(3x)}^{2} – {(3x)}^{3} &- = 125 + 75x – 45x^2 – 27x^3\end{align}

c)\begin{align}{(4n + m^{2})}^{2}\cdot{(4n – m)} \\ =  {(4n)}^{3} + {(4n)}^{2}\cdot{(m^2)} – 4n\cdot {(m^2)}^2 – {(m^2)}^{3} & = 64n^3 + 16n^2m^2 – 4nm^4 – m^6 \end{align}

d)\begin{align}{(5a + 3b)}^{2 }\cdot{(5a – 3b)} \\ = {(5a)}^{3} +{(5a)}^{2}\cdot {(3b)} – 5a\cdot{(3b)^{2}} – {(3b)}^3 & = 125a^3 + 75a^2b – 45ab^2 – 27b^3\end{align}

e)\begin{align}{(7x + 2y)}^{2}\cdot{7x -2y)} \\ = {(7x)}^{3} + {(7x)}^{2}\cdot {(2y)} – 7x\cdot{(2y)^2} – {(2y)^3} & = 343x^3 + 98x^2y – 14xy^2 – 8y^3 \end{align}

f)\begin{align}{(10 + 3v)}^{2}\cdot{(10 – 3v)} \\ = (10)^3 + (10)^2\cdot {(3v)} – 10\cdot{(3v)}^2 – {(3v)}^3 & = 1000 + 300v – 90v^2 – 9v^3\end{align}

g)\begin{align}{(px + qy)}^{2}\cdot{(px – qy)} \\ =  {(px)}^3 + {(px)}^2\cdot {(qy)} – px\cdot{(qy)}^2 – {(qy)}^3 & =  p^3x^3 + p^2qx^2y – pq^2xy^2 – q^3y^3\end{align}

7. Agora vamos multiplicar o quadrado das diferenças, pelas somas dos dois números, conforme a regra vista. a)$\color{Sepia}{{(3x – 2y)}^{2}\cdot{(3x + 2y)}}$; b)$\color{Sepia}{{(5a – bx)}^{2}\cdot{(5a + bx)}}$; c)$\color{Sepia}{{(1 – 5x)}^{2}\cdot{(1 + 5x)}}$; d)$\color{Sepia}{{(6t – 4s)}^{2}\cdot{(6t+ 4s)}}$; e)$\color{Sepia}{{(8l – z)}^{2}\cdot{(8l +z)}}$; f)$\color{Sepia}{{(4n – 5m)}^{2}\cdot{(4n +5m)}}$; g)$\color{Sepia}{{(r – pq)}^{2}\cdot{(r + pq)}}$.

a)\begin{align}{(3x – 2y)}^{2}\cdot{(3x + 2y)} \\  =  {(3x)}^3 – {(3x)}^2\cdot{(2y)} – 3x\cdot {(2y)^2} + {(2y)}^3 & = 27x^3 – 18x^2y – 12xy^2 + 8y^3\end{align}

b)\begin{align}{(5a – bx)}^{2}\cdot{(5a + bx)} \\ = {(5a)}^3 – {(5a)^2}\cdot{(bx)} – 5a\cdot{(bx)}^2 + {(bx)}^3 & = 125 a^3 – 25abx – 5ab^2x^2 + b^3x^3\end{align}

c)\begin{align}{(1 – 5x)}^{2}\cdot{(1 + 5x)} \\ =  1^3 – 1^2\cdot 5x – 1\cdot {(5x)^2} + {(5x)}^3 & =1 – 5x – 25x^2 + 125x^3\end{align}

d)\begin{align}{(6t – 4s)}^{2}\cdot{(6t+ 4s)} \\ = {(6t)}^3 – {(6t)}^2\cdot {(4s)} – 6t\cdot {(4s)}^2 + {(4s)}^3 & =  216t^3 – 144t^2s – 96ts^2 + 64s^3\end{align}

e)\begin{align}{(8i – z)}^{2}\cdot{(8i +z)} \\  = {(8i)}^3 – {(8i)^2}\cdot (z) – 8i\cdot z^2 + z^3 & = 512i^3 – 64i^2z – 8iz^2 + z^3\end{align}

f)\begin{align}{(4n – 5m)}^{2}\cdot{(4n +5m)} \\ =  {(4n)}^3 – {(4n)^2}\cdot{(5m} -4n\cdot {((5m)}^2 + {(5m)}^3 & =  64n^3 – 80mn^2 – 100m^2n + 125m^3\end{align}

g)\begin{align}{(r – pq)}^{2}\cdot{(r + pq)} \\  =  r^3 – r^2\cdot {(pq)} – r\cdot {(pq)}^2 + {(pq)}^3 & =   r^3 – pqr^2 – p^2q^2r + p^3q^3\end{align}

Curitiba, 18 de abril de 2016. Republicado em 21 de dezembro de 2017.

Décio Adams

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