a) x² + 5x + 6

= x² + 2x + 3x + 6

= (x² + 2x) + (3x + 6)

= x(x + 2) + 3 (x + 2)

= (x + 2) (x + 3)

b) x² + 6x + 8

= x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 2)

= (x + 2)(x + 4)

c) x² - 5x - 14

= x² + 2x - 7x - 14

= (x² + 2x) - (7x + 14)

= x(x + 2) - 7(x + 2)

= (x + 2)(x - 7)

d) x² - 9x + 18

= x² - 3x - 6x + 18

= (x² - 3x) - (6x + 18)

= x(x - 3) - 6 (x - 3)

= (x - 3)(x - 6)

e) x² - 7x + 12

= x² -3x - 4x + 12

= (x² - 3x) - (4x + 12)

= x(x - 3) - 4(x - 3)

= (x - 3)(x - 4)

f) 3x² + 9x - 30

= 3(x² + 3x - 10)

= 3\(\left[\left(x^2+5x-2x-10\right)\right]\)

= 3\(\left[\left(x^2+5x\right)-\left(2x-10\right)\right]\)

= 3\(\left[x\left(x+5\right)-2\left(x+5\right)\right]\)

= 3(x + 5)(x - 2)

 Chuc ban hoc tot

a) \(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)

b) \(x^2+6x+8=\left(x+2\right)\left(x+4\right)\)

c) \(x^2-5x-14=\left(x-7\right)\left(x+2\right)\)

d) \(x^2-9x+18=\left(x-3\right)\left(x-6\right)\)

e) \(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)

f) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)

b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)

c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)

d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)

\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)

\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)

\(=3\left(x-2\right)\left(x+5\right)\)

c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)

\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)

\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

a,\(x^2-5x+6=x^2-2x-3x+6 =x(x-2)-3(x-2) =(x-2)(x-3)\)

b,\(3x^2+9x-30=3x^2+15x-6x-30=3x(x+5)-6(x+5)=(x+5)(3x-6)\)

c,\(x^2-7x+12=x^2-4x-3x+12=x(x-4)-3(x-4)=(x-4)(x-3)\)

d,\(x^2-7x+10=x^2-2x-5x+10=x(x-2)-5(x-2)=(x-2)(x-5)\)

Hok tốt !!!!!!!!!!

\(\text{a) }x^2-9x+20\)

\(=x^2-4x-5x+20\)

\(=\left(x^2-4x\right)-\left(5x-20\right)\)

\(=x\left(x-4\right)-5\left(x-4\right)\)

\(=\left(x-4\right)\left(x-5\right)\)

\(\text{b) }x^2+9x+20\)

\(=x^2+4x+5x+20\)

\(=\left(x^2+4x\right)+\left(5x+20\right)\)

\(=x\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(x+5\right)\)

\(\text{c) }x^2+x-20\)

\(=x^2+5x-4x-20\)

\(=\left(x^2+5x\right)-\left(4x+20\right)\)

\(=x\left(x+5\right)-4\left(x+5\right)\)

\(=\left(x+5\right)\left(x-4\right)\)

\(\text{d) }x^2-x-20\)

\(=x^2+4x-5x-20\)

\(=\left(x^2+4x\right)-\left(5x+20\right)\)

\(=x\left(x+4\right)-5\left(x+4\right)\)

\(=\left(x+4\right)\left(x-5\right)\)

a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)

b.  \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)

c.  \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)

a)3x²+9x-30=3x²+15x-6x-30=(3x²-6x)+(15x-30)=3x(x-2)+15(x-2)=(3x+15)(x-2)

b)3x²-5x-2=3x²-6x+x-2=(3x²-6x)+(x-2)=3x(x-2)+(x-2)=(3x+1)(x-2)

a) \(4x^2+4x-3\)

\(=4x^2-2x+6x-3\)

\(=2x\left(2x-1\right)+3\left(2x-1\right)\)

\(=\left(2x+3\right)\left(2x-1\right)\)

b) \(2x^2+xy-y^2\)

\(=2x^2+2xy-xy-y^2\)

\(=2x\left(x+y\right)-y\left(x+y\right)\)

\(=\left(2x-y\right)\left(x+y\right)\)

c) \(x^2+9x-22\)

\(=x^2+11x-2x-22\)

\(=x\left(x+11\right)-2\left(x+11\right)\)

\(=\left(x-2\right)\left(x+11\right)\)

Bài 1 : Phân tích các đa thức sau thành nhân tử : ( tách một hạn tử thành nhiều hạng tử )
a, 3x² + 9x - 30

= 3(x² + 3x - 10)

= 3(x² + 5x - 2x - 10)

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

= 3(x + 5)(x - 2)

b, x² - 3x + 2

= x² - x - 2x + 2

= x(x - 1) - 2(x - 1)

= (x - 1)(x - 2)
c, x² - 9x + 18

= x² - 6x - 3x + 18

= x(x - 6) - 3(x - 6)

= (x - 6)(x - 3)
d, x² - 6x + 8

= x² - 4x - 2x + 8

= x(x - 4) - 2(x - 4)

= (x - 4)(x - 2)
e, x² - 5x - 14

= x² + 2x - 7x - 14

= x(x + 2) - 7(x + 2)

= (x + 2)(x - 7)
f, x² + 6x + 5

= x² + x + 5x + 5

= x(x + 1) + 5(x + 1)

= (x + 1)(x + 5)
h, x² - 7x + 12

= x² - 3x - 4x + 12

= x(x - 3) - 4(x - 3)

= (x - 3)(x - 4)
i, x² - 7x + 10

= x² - 2x - 5x + 10

= x(x - 2) - 5(x - 2)

= (x - 2)(x - 5)

#Học tốt!