No cálculo algébrico, algumas expressões representadas por produtos de expressões algébricas, aparecem com muita frequência. Pela importância que representam no cálculo algébrico, essas expressões são denominadas Produtos notáveis .
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
{\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\,}
(
4
x
5
y
+
z
)
2
=
16
x
2
25
y
2
+
8
x
z
5
y
+
z
2
{\displaystyle \left({\frac {4x}{5y}}+z\right)^{2}={\frac {16x^{2}}{25y^{2}}}+{\frac {8xz}{5y}}+z^{2}}
(
8
x
+
a
)
2
=
64
x
2
+
16
a
x
+
a
2
{\displaystyle (8x+a)^{2}=64x^{2}+16ax+a^{2}\,}
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
{\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\,}
[ editar | editar código-fonte ] A expressão difere, do quadrado da soma, apenas pelo sinal da segunda parcela:
(
x
−
y
)
2
=
(
x
−
y
)
.
(
x
−
y
)
=
x
2
−
x
y
−
y
x
+
y
2
=
x
2
−
2
x
y
+
y
2
{\displaystyle (x-y)^{2}=(x-y).(x-y)=x^{2}-xy-yx+y^{2}=x^{2}-2xy+y^{2}\,}
(
3
m
4
n
−
p
)
2
=
9
m
2
16
n
2
−
6
m
p
4
n
+
p
2
{\displaystyle \left({\frac {3m}{4n}}-p\right)^{2}={\frac {9m^{2}}{16n^{2}}}-{\frac {6mp}{4n}}+p^{2}}
(
1
−
2
x
)
2
=
1
−
4
x
+
4
x
2
{\displaystyle (1-2x)^{2}=1-4x+4x^{2}\,}
(
a
+
b
)
.
(
a
−
b
)
=
a
2
−
a
b
+
b
a
−
b
2
=
a
2
−
b
2
{\displaystyle (a+b).(a-b)=a^{2}-ab+ba-b^{2}=a^{2}-b^{2}\,}
(
a
2
+
b
3
)
.
(
a
2
−
b
3
)
=
a
4
−
b
6
{\displaystyle (a^{2}+b^{3}).(a^{2}-b^{3})=a^{4}-b^{6}\,}
(
a
x
−
2
)
.
(
a
x
+
2
)
=
a
2
x
2
−
4
{\displaystyle \left({\frac {a}{x}}-2\right).\left({\frac {a}{x}}+2\right)={\frac {a^{2}}{x^{2}}}-4}
(
x
−
y
)
3
=
(
x
−
y
)
.
(
x
−
y
)
.
(
x
−
y
)
{\displaystyle (x-y)^{3}=(x-y).(x-y).(x-y)\,}
=
(
x
−
y
)
.
(
x
−
y
)
2
{\displaystyle =(x-y).(x-y)^{2}\,}
=
(
x
−
y
)
.
(
x
2
−
2
x
y
+
y
2
)
{\displaystyle =(x-y).(x^{2}-2xy+y^{2})\,}
=
x
3
−
2
(
x
2
)
y
+
x
y
2
−
y
x
2
+
2
x
y
2
−
y
3
{\displaystyle =x^{3}-2(x^{2})y+xy^{2}-yx^{2}+2xy^{2}-y^{3}\,}
=
x
3
−
3
x
2
y
+
3
x
y
2
−
y
3
{\displaystyle =x^{3}-3x^{2}y+3xy^{2}-y^{3}\,}
(
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}\,}
Decomposição volumétrica do binômio ao cubo O cubo da soma de dois termos diferente do cubo da diferença apenas pelos sinais
(
x
+
y
)
3
=
x
3
+
3
x
2
y
+
3
x
y
2
+
y
3
{\displaystyle (x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}\,}
(
m
+
3
n
)
3
=
m
3
+
9
m
2
n
+
27
m
n
2
+
27
n
3
{\displaystyle (m+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
+
b
+
c
)
2
=
(
a
+
b
+
c
)
.
(
a
+
b
+
c
)
{\displaystyle (a+b+c)^{2}=(a+b+c).(a+b+c)\,}
(
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}
(
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)\,}
=
x
2
+
4
y
2
+
9
−
4
x
y
−
6
x
+
12
y
{\displaystyle =x^{2}+4y^{2}+9-4xy-6x+12y\,}
Considerando o produto notável
(
x
+
a
)
.
(
x
+
b
)
{\displaystyle (x+a).(x+b)\,}
(
x
+
a
)
.
(
x
+
b
)
=
x
2
+
a
x
+
b
x
+
a
b
{\displaystyle (x+a).(x+b)=x^{2}+ax+bx+ab\,}
⇒
(
x
+
a
)
.
(
x
+
b
)
=
x
2
+
(
a
+
b
)
x
+
a
b
{\displaystyle \Rightarrow (x+a).(x+b)=x^{2}+(a+b)x+ab\,}
(
x
+
4
)
(
x
+
3
)
=
x
2
+
(
4
+
3
)
x
+
4.3
=
x
2
+
7
x
+
12
{\displaystyle (x+4)(x+3)=x^{2}+(4+3)x+4.3=x^{2}+7x+12\,}
(
x
−
2
)
(
x
−
6
)
=
x
2
+
(
−
2
−
6
)
x
+
(
−
2
)
(
−
6
)
=
x
2
−
8
x
+
12
{\displaystyle (x-2)(x-6)=x^{2}+(-2-6)x+(-2)(-6)=x^{2}-8x+12\,}
(
x
−
1
)
(
x
+
5
)
=
x
2
+
(
−
1
+
5
)
x
+
5
(
−
1
)
=
x
2
+
4
x
−
5
{\displaystyle (x-1)(x+5)=x^{2}+(-1+5)x+5(-1)=x^{2}+4x-5\,}
