Développement et factorisation

Utiliser la distributivité et les identités remarquables - Exercice 1

12 min
30
Question 1
COMPETENCES :Savoir calculer et factoriser une expression en utilisant le langage algébrique.

A=(2x+5)(x1)+(3x7)(8x+3)A=\left(2x+5\right)\left(x-1\right)+\left(3x-7\right)\left(8x+3\right)

Correction
A=(2x+5)(x1)+(3x7)(8x+3)A={\color{brown}\left(2x+5\right)\left(x-1\right)}+{\color{green}\left(3x-7\right)\left(8x+3\right)}
A=2x×x+2x×(1)+5×x+5×(1)+3x×8x+3x×3+(7)×8x+(7)×3\small{A={\color{brown}2x\times x+2x\times \left(-1\right)+5\times x+5\times \left(-1\right)}+{\color{green}3x\times 8x+3x\times 3+\left(-7\right)\times 8x+\left(-7\right)\times 3}}
A=2x22x+5x5+24x2+9x56x21A=2x^{2} -2x+5x-5+24x^{2} +9x-56x-21
A=2x2+24x22x+5x+9x56x215A=2x^{2} +24x^{2} -2x+5x+9x-56x-21-5
A=26x244x26\color{blue}A=26x^{2} -44x-26
Question 2

B=(x+5)2+7B =(x+5)^2+7

Correction
  • (a+b)2=a2+2ab+b2\left({\color{blue}a}+{\color{red}b}\right)^{2} ={\color{blue}a}^{2} +2{\color{blue}a}{\color{red}b}+{\color{red}b}^{2}

B=(x+5)2+7B={\color{brown}(x+5)^2}+7 équivaut successivement à :
B=(x2+2×x×5+52)+7B={\color{brown}(x^2+2\times{x}\times{5}+5^2)}+7
B=(x2+10x+25)+7B={\color{brown}(x^2+10x+25)}+7
B=x2+10x+25+7B=x^2+10x+25+7
B=x2+10x+32\color{blue}B=x^{2} +10x+32
Question 3

C=(2x+1)(x5)+(5x3)(3x2)C=\left(2x+1\right)\left(x-5\right)+\left(5x-3\right)\left(3x-2\right)

Correction
C=(2x+1)(x5)+(5x3)(3x2)C={\color{brown}\left(2x+1\right)\left(x-5\right)}+{\color{green}\left(5x-3\right)\left(3x-2\right)} équivaut successivement à :
C=2x×x+2x×(5)+1×x+1×(5)+5x×3x+5x×(2)+(3)×3x+(3)×(2)\small{C={\color{brown}2x\times x+2x\times \left(-5\right)+1\times x+1\times \left(-5\right)}+{\color{green}5x\times 3x+5x\times \left(-2\right)+\left(-3\right)\times 3x+\left(-3\right)\times \left(-2\right)}}
C=2x210x+x5+15x210x9x+6C=2x^{2} -10x+x-5+15x^{2} -10x-9x+6
C=17x228x+1\color{blue}C=17x^{2} -28x+1