résolu

Bonjour, j'aurais besoin d'aide :) merci

Factoriser:

a) [tex]2a+12[/tex]
b) [tex]5a^{2}+8a[/tex]
c) [tex]2a^{2}+8a[/tex]
d) [tex]2t (t+3)-5t[/tex]
e) [tex](y+2)(y-7)+16(y-7)[/tex]
f) [tex](2t-5) (3t+2) + 8t(2t-5)[/tex]
g) [tex](5x+2) (2x+3) - (5x+2)^{2} [/tex]
h) [tex]2(y-5)^{2} - 3(y-5)[/tex]
i) [tex]3(x-1)^{2} + x-1[/tex]

Répondre :

tommus
a) 2 (a + 6)

b) a (5a + 8)

c) 2a (a + 4)

d) 2t (t + 3) - 5t = 2t² + 6t - 5t = 2t² + t = t (2t + 1)

e) (y - 7) [(y + 2) + 16] = (y - 7) (y + 18)

f) (2t - 5) (3t + 2) + 8t (2t - 5) = (2t - 5) [(3t + 2) + 8t] = (2t - 5) (11t + 2)

g) (5x + 2) (2x + 3) - (5x + 2)² = (5x + 2) [(2x + 3) - (5x + 2)] = (5x + 2) (2x + 3 - 5x - 2) = (5x + 2) (-3x + 1)

h) 2 (y - 5)² - 3 (y - 5) = (y - 5) [2 (y - 5) - 3] = (y - 5) (2y - 13)

i) 3 (x - 1)² + x - 1 = 3 (x - 1)² + (x - 1) = (x - 1) [3(x - 1) + 1) = (x - 1) (3x - 2)
a) 2(a+6)
b) a(5a+8)
c) 2a(a+4)
d) 2t (t+3)-5t
donc t( 2(t+3) -5)
donc t(2t+6-5)
donc t(2t+1)
e) (y+2)(y-7)+16(y-7)
donc (y-7) ( (y+2) + 16 )
donc (y-7)(y+18)
f) (2t-5) (3t+2) + 8t(2t-5)
donc (2t-5) ( (3t+2)+8t )
donc (2t-5)(11t+2)
g) (5x+2) (2x+3) - (5x+2)^{2}
donc (5x+2) ( (2x+3) -(5x+2) )
donc (5x+2) ( 2x+3-5x-2 )
donc (5x+2)(-3x+1)
h) 2(y-5)^{2} - 3(y-5)
donc (y-5) ( 2(y-5) - 3 )
donc (y-5) ( 2y-10-3)
donc (y-5)(2y-13)
 i) 3(x-1)^{2} + x-1
(x-1)( 3(x-1) +1 )
donc (x-1)(3x-3+1) donc (x-1)(3x-2)

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