En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : FactorisationIdentités remarquables/Exercices/Factorisation », n'a pu être restituée correctement ci-dessus.
Factoriser en utilisant les identités remarquables.
a)
x
2
+
14
x
+
49
{\displaystyle x^{2}+14x+49}
b)
9
x
2
−
6
x
+
1
{\displaystyle 9x^{2}-6x+1}
c)
2
x
2
+
x
24
+
3
{\displaystyle 2x^{2}+x{\sqrt {24}}+3}
d)
1
4
−
3
x
+
9
x
2
{\displaystyle {\frac {1}{4}}-3x+9x^{2}}
e)
x
2
−
25
{\displaystyle x^{2}-25}
f)
9
x
2
−
16
{\displaystyle 9x^{2}-16}
g)
3
x
2
−
1
{\displaystyle 3x^{2}-1}
h)
(
x
−
3
)
2
−
16
{\displaystyle (x-3)^{2}-16}
i)
(
x
−
3
)
2
−
2
{\displaystyle (x-3)^{2}-2}
j)
4
(
x
−
3
)
2
−
25
(
1
−
3
x
)
2
{\displaystyle 4(x-3)^{2}-25(1-3x)^{2}}
Solution
a)
x
2
+
14
x
+
49
=
x
2
+
2
×
7
x
+
7
2
=
(
x
+
7
)
2
{\displaystyle x^{2}+14x+49=x^{2}+2\times 7x+7^{2}=(x+7)^{2}}
b)
9
x
2
−
6
x
+
1
=
(
3
x
)
2
−
2
×
3
x
+
1
2
=
(
3
x
+
1
)
2
{\displaystyle 9x^{2}-6x+1=(3x)^{2}-2\times 3x+1^{2}=(3x+1)^{2}}
c)
2
x
2
+
x
24
+
3
=
2
x
2
+
2
x
6
+
3
=
(
x
2
)
2
+
2
×
x
2
×
3
+
(
3
)
2
=
(
x
2
+
3
)
2
{\displaystyle 2x^{2}+x{\sqrt {24}}+3=2x^{2}+2x{\sqrt {6}}+3=(x{\sqrt {2}})^{2}+2\times x{\sqrt {2}}\times {\sqrt {3}}+({\sqrt {3}})^{2}=(x{\sqrt {2}}+{\sqrt {3}})^{2}}
d)
1
4
−
3
x
+
9
x
2
=
(
1
2
)
2
−
2
×
1
2
3
x
+
(
3
x
)
2
=
(
1
2
−
3
x
)
2
{\displaystyle {\frac {1}{4}}-3x+9x^{2}=\left({\frac {1}{2}}\right)^{2}-2\times {\frac {1}{2}}3x+(3x)^{2}=\left({\frac {1}{2}}-3x\right)^{2}}
e)
x
2
−
25
=
x
2
−
5
2
=
(
x
+
5
)
(
x
−
5
)
{\displaystyle x^{2}-25=x^{2}-5^{2}=(x+5)(x-5)}
f)
9
x
2
−
16
=
(
3
x
)
2
−
4
2
=
(
3
x
+
4
)
(
3
x
−
4
)
{\displaystyle 9x^{2}-16=(3x)^{2}-4^{2}=(3x+4)(3x-4)}
g)
3
x
2
−
1
=
(
x
3
)
2
−
1
2
=
(
x
3
+
1
)
(
x
3
−
1
)
{\displaystyle 3x^{2}-1=(x{\sqrt {3}})^{2}-1^{2}=(x{\sqrt {3}}+1)(x{\sqrt {3}}-1)}
h)
(
x
−
3
)
2
−
16
=
(
x
−
3
)
2
−
4
2
=
(
x
−
3
+
4
)
(
x
−
3
−
4
)
=
(
x
+
1
)
(
x
−
7
)
{\displaystyle (x-3)^{2}-16=(x-3)^{2}-4^{2}=(x-3+4)(x-3-4)=(x+1)(x-7)}
i)
(
x
−
3
)
2
−
2
=
(
x
−
3
)
2
−
(
2
)
2
=
(
x
−
3
+
2
)
(
x
−
3
−
2
)
{\displaystyle (x-3)^{2}-2=(x-3)^{2}-({\sqrt {2}})^{2}=(x-3+{\sqrt {2}})(x-3-{\sqrt {2}})}
j)
4
(
x
−
3
)
2
−
25
(
1
−
3
x
)
2
=
[
2
(
x
−
3
)
]
2
−
[
5
(
1
−
3
x
)
]
2
=
[
2
(
x
−
3
)
+
5
(
1
−
3
x
)
]
[
2
(
x
−
3
)
−
5
(
1
−
3
x
)
]
=
(
2
x
−
6
+
5
−
15
x
)
(
2
x
−
6
−
5
+
15
x
)
=
(
−
13
x
−
1
)
(
17
x
−
11
)
{\displaystyle 4(x-3)^{2}-25(1-3x)^{2}=[2(x-3)]^{2}-[5(1-3x)]^{2}=[2(x-3)+5(1-3x)][2(x-3)-5(1-3x)]=(2x-6+5-15x)(2x-6-5+15x)=(-13x-1)(17x-11)}
Factoriser les expressions suivantes :
a)
(
3
x
+
1
)
(
−
4
x
+
5
)
−
2
(
3
x
+
1
)
{\displaystyle (3x+1)(-4x+5)-2(3x+1)}
b)
(
x
−
5
)
(
x
+
3
)
−
(
2
x
+
7
)
(
x
+
3
)
{\displaystyle (x-5)(x+3)-(2x+7)(x+3)}
c)
(
x
+
2
)
(
5
x
−
1
)
−
(
5
x
+
2
)
(
2
x
+
4
)
{\displaystyle (x+2)(5x-1)-(5x+2)(2x+4)}
d)
(
2
x
+
3
)
2
−
(
4
x
+
6
)
(
−
4
x
+
2
)
{\displaystyle (2x+3)^{2}-(4x+6)(-4x+2)}
Solution
a)
(
3
x
+
1
)
(
−
4
x
+
5
)
−
2
(
3
x
+
1
)
=
(
3
x
+
1
)
[
(
−
4
x
+
5
)
−
2
]
=
(
3
x
+
1
)
(
−
4
x
+
3
)
{\displaystyle (3x+1)(-4x+5)-2(3x+1)=(3x+1)[(-4x+5)-2]=(3x+1)(-4x+3)}
b)
(
x
−
5
)
(
x
+
3
)
−
(
2
x
+
7
)
(
x
+
3
)
=
(
x
+
3
)
[
(
x
−
5
)
−
(
2
x
+
7
)
]
=
(
x
+
3
)
(
−
x
−
12
)
{\displaystyle (x-5)(x+3)-(2x+7)(x+3)=(x+3)[(x-5)-(2x+7)]=(x+3)(-x-12)}
c)
(
x
+
2
)
(
5
x
−
1
)
−
(
5
x
+
2
)
(
2
x
+
4
)
=
(
x
+
2
)
(
5
x
−
1
)
−
2
(
5
x
+
2
)
(
x
+
2
)
=
(
x
+
2
)
[
(
5
x
−
1
)
−
2
(
5
x
+
2
)
]
=
(
x
+
2
)
(
5
x
−
1
−
10
x
−
4
)
=
(
x
+
2
)
(
−
5
x
−
5
)
=
−
5
(
x
+
2
)
(
x
+
1
)
{\displaystyle (x+2)(5x-1)-(5x+2)(2x+4)=(x+2)(5x-1)-2(5x+2)(x+2)=(x+2)[(5x-1)-2(5x+2)]=(x+2)(5x-1-10x-4)=(x+2)(-5x-5)=-5(x+2)(x+1)}
d)
(
2
x
+
3
)
2
−
(
4
x
+
6
)
(
−
4
x
+
2
)
=
(
2
x
+
3
)
2
−
2
(
2
x
+
3
)
(
−
4
x
+
2
)
=
(
2
x
+
3
)
[
(
2
x
+
3
)
−
2
(
−
4
x
+
2
)
]
=
(
2
x
+
3
)
(
2
x
+
3
+
8
x
−
4
)
=
(
2
x
+
3
)
(
10
x
−
1
)
{\displaystyle (2x+3)^{2}-(4x+6)(-4x+2)=(2x+3)^{2}-2(2x+3)(-4x+2)=(2x+3)[(2x+3)-2(-4x+2)]=(2x+3)(2x+3+8x-4)=(2x+3)(10x-1)}
Factoriser les expressions suivantes :
a)
4
x
2
−
1
−
(
2
x
−
1
)
(
3
x
+
5
)
{\displaystyle 4x^{2}-1-(2x-1)(3x+5)}
b)
4
x
2
+
4
x
+
1
−
(
2
x
+
1
)
(
3
x
+
5
)
{\displaystyle 4x^{2}+4x+1-(2x+1)(3x+5)}
c)
a
b
(
a
+
b
)
+
b
c
(
a
2
+
2
a
b
+
b
2
)
+
b
2
(
a
2
−
b
2
)
{\displaystyle ab(a+b)+bc(a^{2}+2ab+b^{2})+b^{2}(a^{2}-b^{2})}
d)
3
x
(
x
−
1
)
−
6
(
x
2
−
2
x
+
1
)
+
9
x
(
x
2
−
1
)
{\displaystyle 3x(x-1)-6(x^{2}-2x+1)+9x(x^{2}-1)}
e)
2
x
y
(
y
2
−
9
)
−
4
x
(
3
−
y
)
+
8
x
(
y
2
−
6
y
+
9
)
{\displaystyle 2xy(y^{2}-9)-4x(3-y)+8x(y^{2}-6y+9)}
Solution
a
)
4
x
2
−
1
−
(
2
x
−
1
)
(
3
x
+
5
)
=
(
2
x
)
2
−
1
−
(
2
x
−
1
)
(
3
x
+
5
)
=
(
2
x
+
1
)
(
2
x
−
1
)
−
(
2
x
−
1
)
(
3
x
+
5
)
=
(
2
x
−
1
)
[
(
2
x
+
1
)
−
(
3
x
+
5
)
]
=
(
2
x
−
1
)
(
−
x
−
4
)
=
−
(
2
x
−
1
)
(
x
+
4
)
{\displaystyle {\begin{aligned}a)\quad 4x^{2}-1-(2x-1)(3x+5)&=(2x)^{2}-1-(2x-1)(3x+5)\\&=(2x+1)(2x-1)-(2x-1)(3x+5)\\&=(2x-1)[(2x+1)-(3x+5)]\\&=(2x-1)(-x-4)\\&=-(2x-1)(x+4)\end{aligned}}}
b
)
4
x
2
+
4
x
+
1
−
(
2
x
+
1
)
(
3
x
+
5
)
=
(
2
x
)
2
+
2
×
2
x
+
1
−
(
2
x
+
1
)
(
3
x
+
5
)
=
(
2
x
+
1
)
2
−
(
2
x
+
1
)
(
3
x
+
5
)
=
(
2
x
+
1
)
[
(
2
x
+
1
)
−
(
3
x
+
5
)
]
=
(
2
x
+
1
)
(
−
x
−
4
)
=
−
(
2
x
+
1
)
(
x
+
4
)
{\displaystyle {\begin{aligned}b)\quad 4x^{2}+4x+1-(2x+1)(3x+5)&=(2x)^{2}+2\times 2x+1-(2x+1)(3x+5)\\&=(2x+1)^{2}-(2x+1)(3x+5)\\&=(2x+1)[(2x+1)-(3x+5)]\\&=(2x+1)(-x-4)\\&=-(2x+1)(x+4)\end{aligned}}}
c
)
a
b
(
a
+
b
)
+
b
c
(
a
2
+
2
a
b
+
b
2
)
+
b
2
(
a
2
−
b
2
)
=
a
b
(
a
+
b
)
+
b
c
(
a
+
b
)
2
+
b
2
(
a
+
b
)
(
a
−
b
)
=
b
(
a
+
b
)
[
a
+
c
(
a
+
b
)
+
b
(
a
−
b
)
]
=
b
(
a
+
b
)
(
a
+
a
c
+
b
c
+
a
b
−
b
2
)
{\displaystyle {\begin{aligned}c)\quad ab(a+b)+bc(a^{2}+2ab+b^{2})+b^{2}(a^{2}-b^{2})&=ab(a+b)+bc(a+b)^{2}+b^{2}(a+b)(a-b)\\&=b(a+b)[a+c(a+b)+b(a-b)]\\&=b(a+b)(a+ac+bc+ab-b^{2})\end{aligned}}}
d
)
3
x
(
x
−
1
)
−
6
(
x
2
−
2
x
+
1
)
+
9
x
(
x
2
−
1
)
=
3
x
(
x
−
1
)
−
6
(
x
−
1
)
2
+
9
x
(
x
+
1
)
(
x
−
1
)
=
3
(
x
−
1
)
[
x
−
2
(
x
−
1
)
+
3
x
(
x
+
1
)
]
=
3
(
x
−
1
)
(
3
x
2
+
2
x
+
2
)
{\displaystyle {\begin{aligned}d)\quad 3x(x-1)-6(x^{2}-2x+1)+9x(x^{2}-1)&=3x(x-1)-6(x-1)^{2}+9x(x+1)(x-1)\\&=3(x-1)[x-2(x-1)+3x(x+1)]\\&=3(x-1)(3x^{2}+2x+2)\end{aligned}}}
e
)
2
x
y
(
y
2
−
9
)
−
4
x
(
3
−
y
)
+
8
x
(
y
2
−
6
y
+
9
)
=
2
x
y
(
y
+
3
)
(
y
−
3
)
+
4
x
(
y
−
3
)
+
8
x
(
y
−
3
)
2
=
2
x
(
y
−
3
)
[
y
(
y
+
3
)
+
2
+
4
(
y
−
3
)
]
=
2
x
(
y
−
3
)
[
y
2
+
3
y
+
2
+
4
y
−
12
]
=
2
x
(
y
−
3
)
(
y
2
+
7
y
−
10
)
{\displaystyle {\begin{aligned}e)\quad 2xy(y^{2}-9)-4x(3-y)+8x(y^{2}-6y+9)&=2xy(y+3)(y-3)+4x(y-3)+8x(y-3)^{2}\\&=2x(y-3)[y(y+3)+2+4(y-3)]\\&=2x(y-3)[y^{2}+3y+2+4y-12]\\&=2x(y-3)(y^{2}+7y-10)\end{aligned}}}
Factoriser les expressions suivantes :
a)
(
x
+
3
y
)
2
−
(
x
−
5
y
)
2
{\displaystyle (x+3y)^{2}-(x-5y)^{2}}
b)
(
x
2
+
2
x
y
+
y
2
)
−
(
m
2
−
2
m
n
+
n
2
)
{\displaystyle (x^{2}+2xy+y^{2})-(m^{2}-2mn+n^{2})}
c)
(
5
+
2
5
a
+
a
)
−
(
3
−
2
3
b
+
b
)
{\displaystyle (5+2{\sqrt {5a}}+a)-(3-2{\sqrt {3b}}+b)}
d)
a
2
−
b
2
−
c
2
−
2
b
c
{\displaystyle a^{2}-b^{2}-c^{2}-2bc}
Solution
a
)
(
x
+
3
y
)
2
−
(
x
−
5
y
)
2
=
(
x
+
3
y
+
x
−
5
y
)
(
x
+
3
y
−
x
+
5
y
)
=
(
2
x
−
2
y
)
(
8
y
)
=
16
y
(
x
−
y
)
{\displaystyle {\begin{aligned}a)\quad (x+3y)^{2}-(x-5y)^{2}&=(x+3y+x-5y)(x+3y-x+5y)\\&=(2x-2y)(8y)\\&=16y(x-y)\end{aligned}}}
b
)
(
x
2
+
2
x
y
+
y
2
)
−
(
m
2
−
2
m
n
+
n
2
)
=
(
x
+
y
)
2
−
(
m
−
n
)
2
=
(
x
+
y
+
m
−
n
)
(
x
+
y
−
m
+
n
)
{\displaystyle {\begin{aligned}b)\quad (x^{2}+2xy+y^{2})-(m^{2}-2mn+n^{2})&=(x+y)^{2}-(m-n)^{2}\\&=(x+y+m-n)(x+y-m+n)\end{aligned}}}
c
)
(
5
+
2
5
a
+
a
)
−
(
3
−
2
3
b
+
b
)
=
(
(
5
)
2
+
2
5
a
+
(
a
)
2
)
−
(
(
3
)
2
−
2
3
b
+
(
b
)
2
)
=
(
5
+
a
)
2
−
(
3
−
b
)
2
=
(
5
+
a
+
3
−
b
)
(
5
+
a
−
3
+
b
)
{\displaystyle {\begin{aligned}c)\quad (5+2{\sqrt {5a}}+a)-(3-2{\sqrt {3b}}+b)&=(({\sqrt {5}})^{2}+2{\sqrt {5}}{\sqrt {a}}+({\sqrt {a}})^{2})-(({\sqrt {3}})^{2}-2{\sqrt {3}}{\sqrt {b}}+({\sqrt {b}})^{2})\\&=({\sqrt {5}}+{\sqrt {a}})^{2}-({\sqrt {3}}-{\sqrt {b}})^{2}\\&=({\sqrt {5}}+{\sqrt {a}}+{\sqrt {3}}-{\sqrt {b}})({\sqrt {5}}+{\sqrt {a}}-{\sqrt {3}}+{\sqrt {b}})\end{aligned}}}
d
)
a
2
−
b
2
−
c
2
−
2
b
c
=
a
2
−
(
b
2
+
c
2
+
2
b
c
)
=
a
2
−
(
b
+
c
)
2
=
(
a
+
b
+
c
)
(
a
−
b
−
c
)
{\displaystyle {\begin{aligned}d)\quad a^{2}-b^{2}-c^{2}-2bc&=a^{2}-\left(b^{2}+c^{2}+2bc\right)\\&=a^{2}-\left(b+c\right)^{2}\\&=(a+b+c)(a-b-c)\end{aligned}}}
Factoriser les expressions suivantes :
a)
a
b
c
2
−
a
b
d
2
+
c
e
−
d
e
{\displaystyle abc^{2}-abd^{2}+ce-de}
b)
a
2
+
2
a
b
(
a
−
b
)
−
b
2
{\displaystyle a^{2}+2ab(a-b)-b^{2}}
c)
x
2
+
2
+
1
x
2
{\displaystyle x^{2}+2+{\frac {1}{x^{2}}}}
d)
a
x
+
a
y
+
b
x
+
b
y
+
(
a
−
b
)
2
+
4
a
b
{\displaystyle ax+ay+bx+by+(a-b)^{2}+4ab}
Solution
a
)
a
b
c
2
−
a
b
d
2
+
c
e
−
d
e
=
a
b
(
c
2
−
d
2
)
+
e
(
c
−
d
)
=
a
b
(
c
+
d
)
(
c
−
d
)
+
e
(
c
−
d
)
=
(
c
−
d
)
[
a
b
(
c
+
d
)
+
e
]
=
(
c
−
d
)
(
a
b
c
+
a
b
d
+
e
)
{\displaystyle {\begin{aligned}a)\quad abc^{2}-abd^{2}+ce-de&=ab(c^{2}-d^{2})+e(c-d)\\&=ab(c+d)(c-d)+e(c-d)\\&=(c-d)[ab(c+d)+e]\\&=(c-d)(abc+abd+e)\end{aligned}}}
b
)
a
2
+
2
a
b
(
a
−
b
)
−
b
2
=
a
2
−
b
2
+
2
a
b
(
a
−
b
)
=
(
a
+
b
)
(
a
−
b
)
+
2
a
b
(
a
−
b
)
=
(
a
−
b
)
(
a
+
b
+
2
a
b
)
{\displaystyle {\begin{aligned}b)\quad a^{2}+2ab(a-b)-b^{2}&=a^{2}-b^{2}+2ab(a-b)\\&=(a+b)(a-b)+2ab(a-b)\\&=(a-b)(a+b+2ab)\end{aligned}}}
c
)
x
2
+
2
+
1
x
2
=
x
2
+
2
x
1
x
+
(
1
x
)
2
=
(
x
+
1
x
)
2
{\displaystyle {\begin{aligned}c)\quad x^{2}+2+{\frac {1}{x^{2}}}&=x^{2}+2x{\frac {1}{x}}+\left({\frac {1}{x}}\right)^{2}\\&=\left(x+{\frac {1}{x}}\right)^{2}\end{aligned}}}
d
)
a
x
+
a
y
+
b
x
+
b
y
+
(
a
−
b
)
2
+
4
a
b
=
a
(
x
+
y
)
+
b
(
x
+
y
)
+
a
2
−
2
a
b
+
b
2
+
4
a
b
=
(
a
+
b
)
(
x
+
y
)
+
a
2
+
2
a
b
+
b
2
=
(
a
+
b
)
(
x
+
y
)
+
(
a
+
b
)
2
=
(
a
+
b
)
(
x
+
y
+
a
+
b
)
{\displaystyle {\begin{aligned}d)\quad ax+ay+bx+by+(a-b)^{2}+4ab&=a(x+y)+b(x+y)+a^{2}-2ab+b^{2}+4ab\\&=(a+b)(x+y)+a^{2}+2ab+b^{2}\\&=(a+b)(x+y)+(a+b)^{2}\\&=(a+b)(x+y+a+b)\end{aligned}}}
Factoriser les expressions suivantes :
a)
(
x
+
2
3
−
5
)
2
−
(
x
−
3
3
+
2
)
2
{\displaystyle (x+2{\sqrt {3}}-5)^{2}-(x-3{\sqrt {3}}+2)^{2}}
b)
x
2
+
2
x
(
x
−
1
)
+
(
x
−
1
)
2
{\displaystyle x^{2}+2x(x-1)+(x-1)^{2}}
c)
m
2
+
1
m
2
+
1
{\displaystyle m^{2}+{\frac {1}{m^{2}}}+1}
d)
4
x
2
+
4
x
(
x
+
3
)
+
(
x
+
3
)
2
+
6
(
x
+
1
)
{\displaystyle 4x^{2}+4x(x+3)+(x+3)^{2}+6(x+1)}
Solution
a
)
(
x
+
2
3
−
5
)
2
−
(
x
−
3
3
+
2
)
2
=
(
x
+
2
3
−
5
+
x
−
3
3
+
2
)
(
x
+
2
3
−
5
−
x
+
3
3
−
2
)
=
(
2
x
−
3
−
3
)
(
5
3
−
7
)
{\displaystyle {\begin{aligned}a)\quad (x+2{\sqrt {3}}-5)^{2}-(x-3{\sqrt {3}}+2)^{2}&=(x+2{\sqrt {3}}-5+x-3{\sqrt {3}}+2)(x+2{\sqrt {3}}-5-x+3{\sqrt {3}}-2)\\&=(2x-{\sqrt {3}}-3)(5{\sqrt {3}}-7)\end{aligned}}}
b
)
x
2
+
2
x
(
x
−
1
)
+
(
x
−
1
)
2
=
[
x
+
(
x
−
1
)
]
2
=
(
2
x
−
1
)
2
{\displaystyle {\begin{aligned}b)\quad x^{2}+2x(x-1)+(x-1)^{2}&=[x+(x-1)]^{2}\\&=(2x-1)^{2}\end{aligned}}}
c
)
m
2
+
1
m
2
+
1
=
m
2
+
2
+
1
m
2
−
1
=
m
2
+
2
m
.
1
m
+
(
1
m
)
2
−
1
2
=
(
m
+
1
m
)
2
−
1
2
=
(
m
+
1
m
+
1
)
(
m
+
1
m
−
1
)
{\displaystyle {\begin{aligned}c)\quad m^{2}+{\frac {1}{m^{2}}}+1&=m^{2}+2+{\frac {1}{m^{2}}}-1\\&=m^{2}+2m.{\frac {1}{m}}+\left({\frac {1}{m}}\right)^{2}-1^{2}\\&=(m+{\frac {1}{m}})^{2}-1^{2}\\&=(m+{\frac {1}{m}}+1)(m+{\frac {1}{m}}-1)\end{aligned}}}
d
)
4
x
2
+
4
x
(
x
+
3
)
+
(
x
+
3
)
2
+
6
(
x
+
1
)
=
(
2
x
)
2
+
2
×
2
x
(
x
+
3
)
+
(
x
+
3
)
2
+
6
(
x
+
1
)
=
(
2
x
+
x
+
3
)
2
+
6
(
x
+
1
)
=
(
3
x
+
3
)
2
+
6
(
x
+
1
)
=
9
(
x
+
1
)
2
+
6
(
x
+
1
)
=
(
x
+
1
)
[
9
(
x
+
1
)
+
6
]
=
(
x
+
1
)
(
9
x
+
15
)
=
3
(
x
+
1
)
(
3
x
+
5
)
{\displaystyle {\begin{aligned}d)\quad 4x^{2}+4x(x+3)+(x+3)^{2}+6(x+1)&=(2x)^{2}+2\times 2x(x+3)+(x+3)^{2}+6(x+1)\\&=(2x+x+3)^{2}+6(x+1)\\&=(3x+3)^{2}+6(x+1)\\&=9(x+1)^{2}+6(x+1)\\&=(x+1)[9(x+1)+6]\\&=(x+1)(9x+15)\\&=3(x+1)(3x+5)\end{aligned}}}
Factoriser les expressions suivantes :
a)
x
2
−
8
x
+
15
{\displaystyle x^{2}-8x+15}
b)
x
2
+
2
x
−
35
{\displaystyle x^{2}+2x-35}
c)
x
2
+
x
−
6
{\displaystyle x^{2}+x-6}
d)
x
2
−
3
x
−
40
{\displaystyle x^{2}-3x-40}
e)
x
2
+
6
x
+
7
{\displaystyle x^{2}+6x+7}
Solution
a
)
x
2
−
8
x
+
15
=
x
2
−
8
x
+
16
−
1
=
x
2
−
8
x
+
4
2
−
1
2
=
(
x
−
4
)
2
−
1
2
=
(
x
−
4
+
1
)
(
x
−
4
−
1
)
=
(
x
−
3
)
(
x
−
5
)
{\displaystyle {\begin{aligned}a)\quad x^{2}-8x+15&=x^{2}-8x+16-1\\&=x^{2}-8x+4^{2}-1^{2}\\&=(x-4)^{2}-1^{2}\\&=(x-4+1)(x-4-1)\\&=(x-3)(x-5)\end{aligned}}}
b
)
x
2
+
2
x
−
35
=
x
2
+
2
x
+
1
2
−
6
2
=
(
x
+
1
)
2
−
6
2
=
(
x
+
1
+
6
)
(
x
+
1
−
6
)
=
(
x
+
7
)
(
x
−
5
)
{\displaystyle {\begin{aligned}b)\quad x^{2}+2x-35&=x^{2}+2x+1^{2}-6^{2}\\&=(x+1)^{2}-6^{2}\\&=(x+1+6)(x+1-6)\\&=(x+7)(x-5)\end{aligned}}}
c
)
x
2
+
x
−
6
=
(
x
+
1
2
)
2
−
1
4
−
6
=
(
x
+
1
2
)
2
−
25
4
=
(
x
+
1
2
)
2
−
(
5
2
)
2
=
(
x
+
1
2
+
5
2
)
(
x
+
1
2
−
5
2
)
=
(
x
+
3
)
(
x
−
2
)
{\displaystyle {\begin{aligned}c)\quad x^{2}+x-6&=\left(x+{\frac {1}{2}}\right)^{2}-{\frac {1}{4}}-6\\&=\left(x+{\frac {1}{2}}\right)^{2}-{\frac {25}{4}}\\&=\left(x+{\frac {1}{2}}\right)^{2}-\left({\frac {5}{2}}\right)^{2}\\&=\left(x+{\frac {1}{2}}+{\frac {5}{2}}\right)\left(x+{\frac {1}{2}}-{\frac {5}{2}}\right)\\&=(x+3)(x-2)\end{aligned}}}
d
)
x
2
−
3
x
−
40
=
(
x
−
3
2
)
2
−
9
4
−
40
=
(
x
−
3
2
)
2
−
(
13
2
)
2
=
(
x
−
3
2
+
13
2
)
(
x
−
3
2
−
13
2
)
=
(
x
+
5
)
(
x
−
8
)
{\displaystyle {\begin{aligned}d)\quad x^{2}-3x-40&=\left(x-{\frac {3}{2}}\right)^{2}-{\frac {9}{4}}-40\\&=\left(x-{\frac {3}{2}}\right)^{2}-\left({\frac {13}{2}}\right)^{2}\\&=\left(x-{\frac {3}{2}}+{\frac {13}{2}}\right)\left(x-{\frac {3}{2}}-{\frac {13}{2}}\right)\\&=(x+5)(x-8)\\\end{aligned}}}
e
)
x
2
+
6
x
+
7
=
(
x
+
3
)
2
−
2
=
(
x
+
3
)
2
−
(
2
)
2
=
(
x
+
3
+
2
)
(
x
+
3
−
2
)
{\displaystyle {\begin{aligned}e)\quad x^{2}+6x+7&=(x+3)^{2}-2\\&=(x+3)^{2}-({\sqrt {2}})^{2}\\&=(x+3+{\sqrt {2}})(x+3-{\sqrt {2}})\end{aligned}}}
Factoriser les expressions suivantes :
a)
3
(
x
2
+
2
x
−
35
)
−
(
x
+
2
)
(
x
−
5
)
{\displaystyle 3(x^{2}+2x-35)-(x+2)(x-5)}
b)
(
3
x
−
1
)
2
−
(
2
x
+
3
)
2
−
(
x
−
4
)
2
{\displaystyle (3x-1)^{2}-(2x+3)^{2}-(x-4)^{2}}
Solution
a
)
3
(
x
2
+
2
x
−
35
)
−
(
x
+
2
)
(
x
−
5
)
=
3
[
(
x
+
1
)
2
−
6
2
]
−
(
x
+
2
)
(
x
−
5
)
=
3
[
(
x
+
1
+
6
)
(
x
+
1
−
6
)
]
−
(
x
+
2
)
(
x
−
5
)
=
3
(
x
+
7
)
(
x
−
5
)
−
(
x
+
2
)
(
x
−
5
)
=
(
x
−
5
)
[
3
(
x
+
7
)
−
(
x
+
2
)
]
=
(
x
−
5
)
(
2
x
+
19
)
{\displaystyle {\begin{aligned}a)\quad 3(x^{2}+2x-35)-(x+2)(x-5)&=3[(x+1)^{2}-6^{2}]-(x+2)(x-5)\\&=3[(x+1+6)(x+1-6)]-(x+2)(x-5)\\&=3(x+7)(x-5)-(x+2)(x-5)\\&=(x-5)[3(x+7)-(x+2)]\\&=(x-5)(2x+19)\end{aligned}}}
b
)
(
3
x
−
1
)
2
−
(
2
x
+
3
)
2
−
(
x
−
4
)
2
=
[
(
3
x
−
1
)
+
(
2
x
+
3
)
]
[
(
3
x
−
1
)
−
(
2
x
+
3
)
]
−
(
x
−
4
)
2
=
(
3
x
−
1
+
2
x
+
3
)
(
3
x
−
1
−
2
x
−
3
)
−
(
x
−
4
)
2
=
(
5
x
+
2
)
(
x
−
4
)
−
(
x
−
4
)
2
=
(
x
−
4
)
[
(
5
x
+
2
)
−
(
x
−
4
)
]
=
(
x
−
4
)
[
4
x
+
6
]
=
2
(
x
−
4
)
(
2
x
+
3
)
{\displaystyle {\begin{aligned}b)\quad (3x-1)^{2}-(2x+3)^{2}-(x-4)^{2}&=[(3x-1)+(2x+3)][(3x-1)-(2x+3)]-(x-4)^{2}\\&=(3x-1+2x+3)(3x-1-2x-3)-(x-4)^{2}\\&=(5x+2)(x-4)-(x-4)^{2}\\&=(x-4)[(5x+2)-(x-4)]\\&=(x-4)[4x+6]\\&=2(x-4)(2x+3)\end{aligned}}}
![{\displaystyle 4(x-3)^{2}-25(1-3x)^{2}=[2(x-3)]^{2}-[5(1-3x)]^{2}=[2(x-3)+5(1-3x)][2(x-3)-5(1-3x)]=(2x-6+5-15x)(2x-6-5+15x)=(-13x-1)(17x-11)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7ff821c18e7e544996ecb4758ab88a82b63bf21)
![{\displaystyle (3x+1)(-4x+5)-2(3x+1)=(3x+1)[(-4x+5)-2]=(3x+1)(-4x+3)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96c427a94bb295cf577b6d83d898d298dc904996)
![{\displaystyle (x-5)(x+3)-(2x+7)(x+3)=(x+3)[(x-5)-(2x+7)]=(x+3)(-x-12)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/122a28c3f627382f8e6fd99289679bf138cb64be)
![{\displaystyle (x+2)(5x-1)-(5x+2)(2x+4)=(x+2)(5x-1)-2(5x+2)(x+2)=(x+2)[(5x-1)-2(5x+2)]=(x+2)(5x-1-10x-4)=(x+2)(-5x-5)=-5(x+2)(x+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9d756b909a5a188dc75806bfd33718935c91d48)
![{\displaystyle (2x+3)^{2}-(4x+6)(-4x+2)=(2x+3)^{2}-2(2x+3)(-4x+2)=(2x+3)[(2x+3)-2(-4x+2)]=(2x+3)(2x+3+8x-4)=(2x+3)(10x-1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7e3e9d95cd569983be5ad3b9a6d83f397913727)
![{\displaystyle {\begin{aligned}a)\quad 4x^{2}-1-(2x-1)(3x+5)&=(2x)^{2}-1-(2x-1)(3x+5)\\&=(2x+1)(2x-1)-(2x-1)(3x+5)\\&=(2x-1)[(2x+1)-(3x+5)]\\&=(2x-1)(-x-4)\\&=-(2x-1)(x+4)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db658027d6f67b7bf8a7e8ab8405b363063b2d65)
![{\displaystyle {\begin{aligned}b)\quad 4x^{2}+4x+1-(2x+1)(3x+5)&=(2x)^{2}+2\times 2x+1-(2x+1)(3x+5)\\&=(2x+1)^{2}-(2x+1)(3x+5)\\&=(2x+1)[(2x+1)-(3x+5)]\\&=(2x+1)(-x-4)\\&=-(2x+1)(x+4)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7405103709f9a9d9c20c02ed6c65e078d00e160)
![{\displaystyle {\begin{aligned}c)\quad ab(a+b)+bc(a^{2}+2ab+b^{2})+b^{2}(a^{2}-b^{2})&=ab(a+b)+bc(a+b)^{2}+b^{2}(a+b)(a-b)\\&=b(a+b)[a+c(a+b)+b(a-b)]\\&=b(a+b)(a+ac+bc+ab-b^{2})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/411eba55983e86b43ba733a19a68f3dffc53c459)
![{\displaystyle {\begin{aligned}d)\quad 3x(x-1)-6(x^{2}-2x+1)+9x(x^{2}-1)&=3x(x-1)-6(x-1)^{2}+9x(x+1)(x-1)\\&=3(x-1)[x-2(x-1)+3x(x+1)]\\&=3(x-1)(3x^{2}+2x+2)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07a1d4967e80c65eedef894a5e62b3b4f56e6b31)
![{\displaystyle {\begin{aligned}e)\quad 2xy(y^{2}-9)-4x(3-y)+8x(y^{2}-6y+9)&=2xy(y+3)(y-3)+4x(y-3)+8x(y-3)^{2}\\&=2x(y-3)[y(y+3)+2+4(y-3)]\\&=2x(y-3)[y^{2}+3y+2+4y-12]\\&=2x(y-3)(y^{2}+7y-10)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba27c731b1a5c7119605d4e90320b97065d868c9)
![{\displaystyle {\begin{aligned}a)\quad abc^{2}-abd^{2}+ce-de&=ab(c^{2}-d^{2})+e(c-d)\\&=ab(c+d)(c-d)+e(c-d)\\&=(c-d)[ab(c+d)+e]\\&=(c-d)(abc+abd+e)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5f1f2e1b30048260962f47d6981f4c38693870d)
![{\displaystyle {\begin{aligned}b)\quad x^{2}+2x(x-1)+(x-1)^{2}&=[x+(x-1)]^{2}\\&=(2x-1)^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e22408eae538a9a98d44b3901ad5f7a7c388d319)
![{\displaystyle {\begin{aligned}d)\quad 4x^{2}+4x(x+3)+(x+3)^{2}+6(x+1)&=(2x)^{2}+2\times 2x(x+3)+(x+3)^{2}+6(x+1)\\&=(2x+x+3)^{2}+6(x+1)\\&=(3x+3)^{2}+6(x+1)\\&=9(x+1)^{2}+6(x+1)\\&=(x+1)[9(x+1)+6]\\&=(x+1)(9x+15)\\&=3(x+1)(3x+5)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de85f17b998c1d45cfa595b617ea0d7c1df4928d)
![{\displaystyle {\begin{aligned}a)\quad 3(x^{2}+2x-35)-(x+2)(x-5)&=3[(x+1)^{2}-6^{2}]-(x+2)(x-5)\\&=3[(x+1+6)(x+1-6)]-(x+2)(x-5)\\&=3(x+7)(x-5)-(x+2)(x-5)\\&=(x-5)[3(x+7)-(x+2)]\\&=(x-5)(2x+19)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8858175d8d7cf57fce9dbc4100e3935e7fce5179)
![{\displaystyle {\begin{aligned}b)\quad (3x-1)^{2}-(2x+3)^{2}-(x-4)^{2}&=[(3x-1)+(2x+3)][(3x-1)-(2x+3)]-(x-4)^{2}\\&=(3x-1+2x+3)(3x-1-2x-3)-(x-4)^{2}\\&=(5x+2)(x-4)-(x-4)^{2}\\&=(x-4)[(5x+2)-(x-4)]\\&=(x-4)[4x+6]\\&=2(x-4)(2x+3)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f6ad375b739ee551f98d63ec568af82860b1e16)
