Automatismes : calcul numérique et algébrique

Factorisation - Exercice 6

10 min
25
Factoriser les expressions suivantes :
Question 1

A=x2+4x+4A=x^{2} +4x+4

Correction
    Identiteˊ remarquable\purple{\text{Identité remarquable}}
  • a2+2×a×b+b2=(a+b)2{\color{blue}a}^{2} +2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}+{\color{red}b}\right)^{2}
A=x2+4x+4A=x^{2} +4x+4
A=x2+2×x×2+22A={\color{blue}x}^{2} +2\times {\color{blue}x}\times {\color{red}2}+{\color{red}2}^{2}
Ici nous avons a=xa={\color{blue}x} et b=2b={\color{red}2}. Il vient alors que :
A=(x+2)2A=\left({\color{blue}x}+{\color{red}2}\right)^{2}
Question 2

B=16x256x+49B=16x^{2} -56x+49

Correction
    Identiteˊ remarquable\purple{\text{Identité remarquable}}
  • a22×a×b+b2=(ab)2{\color{blue}a}^{2} -2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}-{\color{red}b}\right)^{2}
B=16x256x+49B=16x^{2} -56x+49
B=(4x)22×4x×7+72B=\left({\color{blue}4x}\right)^{2} -2\times{\color{blue}4x}\times {\color{red}7}+{\color{red}7}^{2}
Ici nous avons a=4xa={\color{blue}4x} et b=7b={\color{red}7}. Il vient alors que :
B=(4x7)2B=\left({\color{blue}4x}-{\color{red}7}\right)^{2}
Question 3

C=9x2+6x+1C=9x^{2} +6x+1

Correction
    Identiteˊ remarquable\purple{\text{Identité remarquable}}
  • a2+2×a×b+b2=(a+b)2{\color{blue}a}^{2} +2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}+{\color{red}b}\right)^{2}
C=9x2+6x+1C=9x^{2} +6x+1
C=(3x)2+2×3x×1+12C=({\color{blue}3x})^{2} +2\times {\color{blue}3x}\times {\color{red}1}+{\color{red}1}^{2}
Ici nous avons a=3xa={\color{blue}3x} et b=1b={\color{red}1}. Il vient alors que :
C=(3x+1)2C=\left({\color{blue}3x}+{\color{red}1}\right)^{2}
Question 4

D=25x220x+4D=25x^{2} -20x+4

Correction
    Identiteˊ remarquable\purple{\text{Identité remarquable}}
  • a22×a×b+b2=(ab)2{\color{blue}a}^{2} -2\times{\color{blue}a}\times{\color{red}b}+{\color{red}b}^{2}=\left({\color{blue}a}-{\color{red}b}\right)^{2}
D=25x220x+4D=25x^{2} -20x+4
B=(5x)22×5x×2+22B=\left({\color{blue}5x}\right)^{2} -2\times{\color{blue}5x}\times {\color{red}2}+{\color{red}2}^{2}
Ici nous avons a=5xa={\color{blue}5x} et b=2b={\color{red}2}. Il vient alors que :
B=(5x2)2B=\left({\color{blue}5x}-{\color{red}2}\right)^{2}