Question

D.SAD
4. Descompuneți în factori, scoțând factorul comun:
a) 2x(x + 1) - 5(x + 1);
b) 4(2x + 3) -- x(2x + 3) + (2x + 3)(x + 1);
c) 6x(3x - 7) + 5(3x - 7);
d) 14x(2x – 1) - 7(2x - 1);
e) 10x(x + 6) + 5(x + 6)2;
f) 12(x - 3)2 + 6x(x – 3);
g) 21x(3x – 1) - 7(3x – 1)?;
h) 24x(4x + 3) - 8(4x + 3)?;
i) (2x + 5)[3(x - 2) - 2(x - 4)] + 2(2x + 5).
ber​

Answer 1

$\bf a) ~~2x(x + 1) - 5(x + 1) = (x + 1)\cdot(2x - 5)$

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$\bf b) ~~4(2x + 3) - x(2x + 3) + (2x + 3)(x + 1) =$

$\bf (2x + 3)\cdot (4 - x + x + 1) = 5\cdot (2x + 3)$

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$\bf c)~~ 6x(3x - 7) + 5(3x - 7) = (3x - 7)\cdot(6x + 5)$

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$\bf d)~~14x(2x - 1) - 7(2x - 1) = (2x-1)\cdot(14x-7)$

$\bf (2x-1)\cdot7\cdot(2x-1)=7\cdot(2x-1)^2$

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$\bf e) ~~10x(x + 6) + 5(x + 6)^2 = (x+6)\cdot[10x+5(x+6)]$

$\bf (x+6)\cdot5\cdot(2x+x+6)=5\cdot(x+6)\cdot(3x+6)$

$\bf 5\cdot(x+6)\cdot3\cdot(x+2)= 15\cdot(x+6)\cdot(x+2)$

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$\bf f)~~ 12(x - 3)^2 + 6x(x -3) = 6(x-3)\cdot[2(x-3)+x]=$

$\bf 6(x-3)\cdot(2x-6+x)=6(x-3)\cdot(3x-6)=$

$\bf 6(x-3)\cdot3\cdot(3x-6)=18\cdot(x-3)\cdot(x-2)$

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$\bf g)~~21x(3x - 1) - 7(3x - 1)^2 = 7(3x-1)\cdot[3x-(3x-1)]=$

$\bf 7(3x-1)\cdot(3x-3x+1)= 7\cdot(3x-1)$

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$\bf h) ~~24x(4x + 3) - 8(4x + 3)^2 =$

$\bf 8(4x + 3)\cdot[3x - (4x + 3)] = 8(4x + 3)\cdot(3x - 4x - 3) =$

$\bf 8\cdot(4x + 3)\cdot(-x - 3)$

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$\bf i)~~ (2x+5)[3(x-2)-2(x-4)]+2(2x+5)=$

$\bf (2x+5)\cdot[3(x-2)-2(x-4)+2] =$

$\bf (2x+5)\cdot(3x-6-2x+8+2) = (2x+5)\cdot(x+4)$

$==pav38==$