Foram assinalados vários problemas nesta página ou se(c)ção:
No cálculo algébrico , Produtos Notáveis são produtos de expressões algébricas que representam determinadas expressões que aparecem com muita frequência. São utilizados principalmente para a fatoração de polinômios e evitar erros com sinais.[ 1]
Quadrado da soma de dois termos
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
.
{\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}.}
Regra básica : Quadrado do primeiro termo, somado ao dobro do primeiro termo multiplicado pelo segundo termo, somado ao quadrado do segundo termo.[ 2]
Prova :
(
a
+
b
)
2
=
(
a
+
b
)
⋅ ⋅ -->
(
a
+
b
)
=
a
⋅ ⋅ -->
(
a
+
b
)
+
b
⋅ ⋅ -->
(
a
+
b
)
=
a
2
+
a
b
+
a
b
+
b
2
=
a
2
+
2
a
b
+
b
2
{\displaystyle (a+b)^{2}=(a+b)\cdot (a+b)=a\cdot (a+b)+b\cdot (a+b)=a^{2}+ab+ab+b^{2}=a^{2}+2ab+b^{2}}
Exemplo:
(
4
x
5
y
+
z
)
2
=
(
4
x
5
y
+
z
)
⋅ ⋅ -->
(
4
x
5
y
+
z
)
=
4
x
5
y
⋅ ⋅ -->
(
4
x
5
y
+
z
)
+
z
⋅ ⋅ -->
(
4
x
5
y
+
z
)
=
16
x
2
25
y
2
+
4
x
z
5
y
+
4
x
z
5
y
+
z
2
=
16
x
2
25
y
2
+
8
x
z
5
y
+
z
2
{\displaystyle \left({\dfrac {4x}{5y}}+z\right)^{2}=\left({\dfrac {4x}{5y}}+z\right)\cdot \left({\dfrac {4x}{5y}}+z\right)={\dfrac {4x}{5y}}\cdot \left({\dfrac {4x}{5y}}+z\right)+z\cdot \left({\dfrac {4x}{5y}}+z\right)={\dfrac {16x^{2}}{25y^{2}}}+{\dfrac {4xz}{5y}}+{\dfrac {4xz}{5y}}+z^{2}={\dfrac {16x^{2}}{25y^{2}}}+{\frac {8xz}{5y}}+z^{2}}
(
8
x
+
a
)
2
=
(
8
x
+
a
)
⋅ ⋅ -->
(
8
x
+
a
)
=
8
x
⋅ ⋅ -->
(
8
x
+
a
)
+
a
⋅ ⋅ -->
(
8
x
+
a
)
=
64
x
2
+
8
a
x
+
8
a
x
+
a
2
=
64
x
2
+
16
a
x
+
a
2
{\displaystyle (8x+a)^{2}=(8x+a)\cdot (8x+a)=8x\cdot (8x+a)+a\cdot (8x+a)=64x^{2}+8ax+8ax+a^{2}=64x^{2}+16ax+a^{2}}
Quadrado da diferença de dois termos
(
a
− − -->
b
)
2
=
a
2
− − -->
2
a
b
+
b
2
{\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}}
Regra básica : Quadrado do primeiro termo, subtraído o dobro do produto do primeiro termo pelo segundo termo, somado ao quadrado do segundo termo.
Prova :
(
a
− − -->
b
)
2
=
(
a
− − -->
b
)
⋅ ⋅ -->
(
a
− − -->
b
)
=
a
⋅ ⋅ -->
(
a
− − -->
b
)
− − -->
b
⋅ ⋅ -->
(
a
− − -->
b
)
=
a
2
− − -->
a
b
− − -->
a
b
+
b
2
=
a
2
− − -->
2
a
b
+
b
2
{\displaystyle (a-b)^{2}=(a-b)\cdot (a-b)=a\cdot (a-b)-b\cdot (a-b)=a^{2}-ab-ab+b^{2}=a^{2}-2ab+b^{2}}
Exemplos :
(
3
m
4
n
− − -->
p
)
2
=
9
m
2
16
n
2
− − -->
3
m
p
2
n
+
p
2
{\displaystyle \left({\frac {3m}{4n}}-p\right)^{2}={\frac {9m^{2}}{16n^{2}}}-{\frac {3mp}{2n}}+p^{2}}
(
1
− − -->
2
x
)
2
=
1
− − -->
4
x
+
4
x
2
{\displaystyle (1-2x)^{2}=1-4x+4x^{2}}
Produto da soma pela diferença de dois termos
(
a
+
b
)
⋅ ⋅ -->
(
a
− − -->
b
)
=
a
2
− − -->
b
2
{\displaystyle (a+b)\cdot (a-b)=a^{2}-b^{2}}
Regra básica : Quadrado do primeiro termo subtraído o quadrado do segundo termo.
Prova :
(
a
+
b
)
⋅ ⋅ -->
(
a
− − -->
b
)
=
a
2
− − -->
a
b
+
a
b
− − -->
b
2
=
a
2
+
a
b
⋅ ⋅ -->
(
1
− − -->
1
)
− − -->
b
2
=
a
2
− − -->
b
2
{\displaystyle (a+b)\cdot (a-b)=a^{2}-ab+ab-b^{2}=a^{2}+ab\cdot (1-1)-b^{2}=a^{2}-b^{2}}
Exemplos :
(
a
2
+
b
3
)
⋅ ⋅ -->
(
a
2
− − -->
b
3
)
=
a
4
− − -->
a
2
b
3
+
a
2
b
3
− − -->
b
6
=
a
4
+
a
2
b
3
⋅ ⋅ -->
(
1
− − -->
1
)
− − -->
b
6
=
a
4
− − -->
b
6
{\displaystyle (a^{2}+b^{3})\cdot (a^{2}-b^{3})=a^{4}-a^{2}b^{3}+a^{2}b^{3}-b^{6}=a^{4}+a^{2}b^{3}\cdot (1-1)-b^{6}=a^{4}-b^{6}}
(
a
x
+
2
)
.
(
a
x
− − -->
2
)
=
a
2
x
2
− − -->
2
a
x
+
2
a
x
− − -->
4
=
a
2
x
2
+
2
a
⋅ ⋅ -->
(
1
− − -->
1
)
x
− − -->
4
=
a
2
x
2
− − -->
4
{\displaystyle \left({\frac {a}{x}}+2\right).\left({\frac {a}{x}}-2\right)={\frac {a^{2}}{x^{2}}}-{\frac {2a}{x}}+{\frac {2a}{x}}-4={\frac {a^{2}}{x^{2}}}+{\frac {2a\cdot (1-1)}{x}}-4={\frac {a^{2}}{x^{2}}}-4}
Cubo da soma de dois termos
Decomposição volumétrica do binômio ao cubo
(
a
+
b
)
3
=
a
3
+
3
a
2
b
+
3
a
b
2
+
b
3
{\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}}
Regra básica : O cubo do primeiro termo, somado o triplo do produto do quadrado do primeiro termo pelo segundo termo, somado ao triplo do produto do primeiro termo pelo quadrado do segundo termo, somado ao cubo do segundo termo.
Exemplos :
(
m
+
3
n
)
3
=
m
3
+
3
⋅ ⋅ -->
m
2
⋅ ⋅ -->
3
n
+
3
⋅ ⋅ -->
m
⋅ ⋅ -->
(
3
n
)
2
+
(
3
n
)
3
=
m
3
+
9
m
2
n
+
27
m
n
2
+
27
n
3
{\displaystyle (m+3n)^{3}=m^{3}+3\cdot m^{2}\cdot 3n+3\cdot m\cdot (3n)^{2}+(3n)^{3}=m^{3}+9m^{2}n+27mn^{2}+27n^{3}}
(
x
+
2
)
3
=
x
3
+
6
x
2
+
12
x
+
8
{\displaystyle (x+2)^{3}=x^{3}+6x^{2}+12x+8}
(
a
+
3
b
)
3
=
a
3
+
9
a
2
b
+
27
a
b
2
+
27
b
3
{\displaystyle (a+3b)^{3}=a^{3}+9a^{2}b+27ab^{2}+27b^{3}}
Cubo da diferença de dois termos
(
a
− − -->
b
)
3
=
a
3
− − -->
3
a
2
b
+
3
a
b
2
− − -->
b
3
{\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}}
Regra básica : Para calcular o cubo da diferença faça: O cubo do 1° termo, subtraído o triplo do produto do quadrado do 1° termo pelo segundo termo, somado ao triplo do produto do 1° termo pelo quadrado do 2° termo, subtraído o cubo do 2° termo.
(
a
− − -->
b
)
3
=
(
a
− − -->
b
)
⋅ ⋅ -->
(
a
− − -->
b
)
2
=
(
a
− − -->
b
)
⋅ ⋅ -->
(
a
2
− − -->
2
a
b
+
b
2
)
=
a
3
− − -->
2
a
2
b
+
a
b
2
− − -->
a
2
b
+
2
a
b
2
− − -->
b
3
=
a
3
− − -->
3
a
2
b
+
3
a
b
2
− − -->
b
3
{\displaystyle (a-b)^{3}=(a-b)\cdot (a-b)^{2}=(a-b)\cdot (a^{2}-2ab+b^{2})=a^{3}-2a^{2}b+ab^{2}-a^{2}b+2ab^{2}-b^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}}
Exemplos :
(
b
− − -->
2
c
)
3
=
b
3
− − -->
6
b
2
c
+
12
b
c
2
− − -->
8
c
3
{\displaystyle (b-2c)^{3}=b^{3}-6b^{2}c+12bc^{2}-8c^{3}}
(
x
y
− − -->
a
b
)
3
=
x
3
y
3
− − -->
3
a
x
2
b
y
2
+
3
a
2
x
b
2
y
− − -->
a
3
b
3
{\displaystyle \left({\frac {x}{y}}-{\frac {a}{b}}\right)^{3}={\frac {x^{3}}{y^{3}}}-{\frac {3ax^{2}}{by^{2}}}+{\frac {3a^{2}x}{b^{2}y}}-{\frac {a^{3}}{b^{3}}}}
(
1
− − -->
x
)
3
=
1
− − -->
3
x
+
3
x
2
− − -->
x
3
{\displaystyle (1-x)^{3}=1-3x+3x^{2}-x^{3}}
Quadrado da soma de três termos
(
a
+
b
+
c
)
2
=
a
2
+
b
2
+
c
2
+
2
a
b
+
2
a
c
+
2
b
c
{\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc}
(
a
+
b
+
c
)
2
=
a
2
+
a
b
+
a
c
+
b
2
+
a
b
+
b
c
+
a
c
+
b
c
+
c
2
{\displaystyle (a+b+c)^{2}=a^{2}+ab+ac+b^{2}+ab+bc+ac+bc+c^{2}}
→ → -->
(
a
+
b
+
c
)
2
=
a
2
+
b
2
+
c
2
+
2
a
b
+
2
a
c
+
2
b
c
{\displaystyle \rightarrow (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc}
Exemplos :
(
x
+
y
+
z
)
2
=
x
2
+
y
2
+
z
2
+
2
x
y
+
2
x
z
+
2
y
z
{\displaystyle (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2xz+2yz}
(
x
− − -->
2
y
− − -->
3
)
2
=
x
2
+
(
− − -->
2
y
)
2
+
(
− − -->
3
)
2
+
2
x
(
− − -->
2
y
)
+
2
x
(
− − -->
3
)
+
2
(
− − -->
2
y
)
(
− − -->
3
)
{\displaystyle (x-2y-3)^{2}=x^{2}+(-2y)^{2}+(-3)^{2}+2x(-2y)+2x(-3)+2(-2y)(-3)}
(
5
x
+
4
y
+
15
z
)
2
=
25
x
2
+
16
y
2
+
225
z
2
+
40
x
y
+
75
x
z
+
60
y
z
{\displaystyle (5x+4y+15z)^{2}=25x^{2}+16y^{2}+225z^{2}+40xy+75xz+60yz}
Produto de Stevin (produto de 2 binômios com um termo comum)
(
x
+
a
)
⋅ ⋅ -->
(
x
+
b
)
=
x
2
+
(
a
+
b
)
x
+
a
b
{\displaystyle (x+a)\cdot (x+b)=x^{2}+(a+b)x+ab}
(
x
+
a
)
⋅ ⋅ -->
(
x
+
b
)
=
x
⋅ ⋅ -->
(
x
+
b
)
+
a
⋅ ⋅ -->
(
x
+
b
)
=
x
2
+
b
x
+
a
x
+
a
b
⇒ ⇒ -->
x
2
+
(
a
+
b
)
x
+
a
b
{\textstyle (x+a)\cdot (x+b)=x\cdot (x+b)+a\cdot (x+b)=x^{2}+bx+ax+ab\Rightarrow x^{2}+(a+b)x+ab}
Exemplos :
(
x
+
4
)
⋅ ⋅ -->
(
x
+
3
)
=
x
2
+
(
4
+
3
)
⋅ ⋅ -->
x
+
4
⋅ ⋅ -->
3
=
x
2
+
7
x
+
12
{\displaystyle (x+4)\cdot (x+3)=x^{2}+(4+3)\cdot x+4\cdot 3=x^{2}+7x+12}
(
x
− − -->
2
)
⋅ ⋅ -->
(
x
− − -->
6
)
=
x
2
+
(
− − -->
2
− − -->
6
)
⋅ ⋅ -->
x
+
(
− − -->
2
)
⋅ ⋅ -->
(
− − -->
6
)
=
x
2
− − -->
8
x
+
12
{\displaystyle (x-2)\cdot (x-6)=x^{2}+(-2-6)\cdot x+(-2)\cdot (-6)=x^{2}-8x+12}
(
x
− − -->
1
)
⋅ ⋅ -->
(
x
+
5
)
=
x
2
+
(
− − -->
1
+
5
)
⋅ ⋅ -->
x
+
5
⋅ ⋅ -->
(
− − -->
1
)
=
x
2
+
4
x
− − -->
5
{\displaystyle (x-1)\cdot (x+5)=x^{2}+(-1+5)\cdot x+5\cdot (-1)=x^{2}+4x-5}
Este tipo de produto notável pode ser usado para resolver equações polinomiais .
Assumindo uma equação polinomial de grau 2 podemos escrevê-la como:
a
x
2
+
b
x
+
c
=
0
{\displaystyle ax^{2}+bx+c=0}
ou
x
2
− − -->
x
⋅ ⋅ -->
(
x
1
+
x
2
)
+
x
1
x
2
=
0
{\displaystyle x^{2}-x\cdot (x_{1}+x_{2})+x_{1}x_{2}=0}
Onde a segunda pode ser fatorada como
(
x
− − -->
x
1
)
⋅ ⋅ -->
(
x
− − -->
x
2
)
=
0
{\displaystyle (x-x_{1})\cdot (x-x_{2})=0}
e a primeira, como consequência, será:
a
⋅ ⋅ -->
(
x
− − -->
x
1
)
⋅ ⋅ -->
(
x
− − -->
x
2
)
=
0
{\displaystyle a\cdot (x-x_{1})\cdot (x-x_{2})=0}
Produto de Warring
Caso 1
(
a
+
b
)
⋅ ⋅ -->
(
a
2
− − -->
a
b
+
b
2
)
=
a
3
+
b
3
{\displaystyle (a+b)\cdot (a^{2}-ab+b^{2})=a^{3}+b^{3}}
Prova : Considerando
(
a
+
b
)
⋅ ⋅ -->
(
a
2
− − -->
a
b
+
b
2
)
,
{\displaystyle (a+b)\cdot (a^{2}-ab+b^{2}),}
temos:
(
a
+
b
)
⋅ ⋅ -->
(
a
2
− − -->
a
b
+
b
2
)
=
a
3
− − -->
a
2
b
+
a
b
2
+
a
2
b
− − -->
a
b
2
+
b
3
=
a
3
+
b
3
.
{\displaystyle (a+b)\cdot (a^{2}-ab+b^{2})=a^{3}-a^{2}b+ab^{2}+a^{2}b-ab^{2}+b^{3}=a^{3}+b^{3}.}
(
x
+
5
)
⋅ ⋅ -->
(
x
2
− − -->
5
x
+
25
)
=
x
3
+
5
3
=
x
3
+
125
{\displaystyle (x+5)\cdot (x^{2}-5x+25)=x^{3}+5^{3}=x^{3}+125}
(
2
x
+
3
)
⋅ ⋅ -->
(
4
x
2
− − -->
6
x
+
9
)
=
(
2
x
)
3
+
3
3
=
8
x
3
+
27
{\displaystyle (2x+3)\cdot (4x^{2}-6x+9)=(2x)^{3}+3^{3}=8x^{3}+27}
Caso 2
(
a
− − -->
b
)
⋅ ⋅ -->
(
a
2
+
a
b
+
b
2
)
=
a
3
− − -->
b
3
{\displaystyle (a-b)\cdot (a^{2}+ab+b^{2})=a^{3}-b^{3}}
Prova : Considerando
(
a
− − -->
b
)
⋅ ⋅ -->
(
a
2
+
a
b
+
b
2
)
,
{\displaystyle (a-b)\cdot (a^{2}+ab+b^{2}),}
temos:
(
a
− − -->
b
)
⋅ ⋅ -->
(
a
2
+
a
b
+
b
2
)
=
a
3
+
a
2
b
− − -->
a
b
2
− − -->
a
2
b
+
a
b
2
− − -->
b
3
=
a
3
− − -->
b
3
{\displaystyle (a-b)\cdot (a^{2}+ab+b^{2})=a^{3}+a^{2}b-ab^{2}-a^{2}b+ab^{2}-b^{3}=a^{3}-b^{3}}
Exemplo
(
x
− − -->
3
)
⋅ ⋅ -->
(
x
2
+
3
x
+
9
)
=
x
3
− − -->
3
3
=
x
3
− − -->
27
{\displaystyle (x-3)\cdot (x^{2}+3x+9)=x^{3}-3^{3}=x^{3}-27}
Ver também
Equação polinomial
Notas e referências
Referências
Ligações externas