Question

Phân tích các đa thức sau thành nhân tử:

a) (x + 2). (x + 3). (x + 4). (x + 5) - 24

b) (4x + 1). (12x - 1). (3x + 2). (x + 1) - 4

c) (x² + 3x + 1). (x² + 3x + 2) - 6

d) 4. (x + 5). (x + 6). (x + 10). (x + 12) - 3x²

Answer 1

c)

Đặt $x^2+3x+1=a$. Khi đó:

\((x^2+3x+1)(x^2+3x+2)-6=a(a+1)-6\)

\(=a^2+3a-2a-6=a(a+3)-2(a+3)=(a-2)(a+3)\)

\(=(x^2+3x-1)(x^2+3x+4)\)

d)

\(4(x+5)(x+6)(x+10)(x+12)-3x^2\)

\(=4[(x+5)(x+12)][(x+6)(x+10)]-3x^2\)

\(=4(x^2+17x+60)(x^2+16x+60)-3x^2\)

\(=4(a+x)a-3x^2\) (đặt \(x^2+16x+60=a\))

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

\(=2a(2a-x)+3x(2a-x)=(2a-x)(2a+3x)\)

\(=(2x^2+32x+120-x)(2x^2+32x+120+3x)\)

\(=(2x^2+31x+120)(2x^2+35x+120)\)

\(=[2x(x+8)+15(x+8)](2x^2+35x+120)\)

\(=(2x+15)(x+8)(2x^2+35x+120)\)

Answer 2

a)

\((x+2)(x+3)(x+4)(x+5)-24\)

\(=[[(x+2)(x+5)]][(x+3)(x+4)]-24\)

\(=(x^2+7x+10)(x^2+7x+12)-24\)

\(=a(a+2)-24\) (đặt $x^2+7x+10=a$)

\(=a^2+2a-24=a^2+6a-4a-24\)

\(=a(a+6)-4(a+6)=(a-4)(a+6)\)

\(=(x^2+7x+6)(x^2+7x+16)\)

\(=(x+1)(x+6)(x^2+7x+16)\)

b)

\((4x+1)(12x-1)(3x+2)(x+1)-4\)

\(=[(4x+1)(3x+2)][(12x-1)(x+1)]-4\)

\(=(12x^2+11x+2)(12x^2+11x-1)-4\)

\(=(a+2)(a-1)-4\) (đặt $12x^2+11x=a$)

\(=a^2+a-6=a^2+3a-2a-6=a(a+3)-2(a+3)\)

\(=(a-2)(a+3)=(12x^2+11x-2)(12x^2+11x+3)\)

Answer 3

c)

Đặt $x^2+3x+1=a$. Khi đó:

\((x^2+3x+1)(x^2+3x+2)-6=a(a+1)-6\)

\(=a^2+3a-2a-6=a(a+3)-2(a+3)=(a-2)(a+3)\)

\(=(x^2+3x-1)(x^2+3x+4)\)

d)

\(4(x+5)(x+6)(x+10)(x+12)-3x^2\)

\(=4[(x+5)(x+12)][(x+6)(x+10)]-3x^2\)

\(=4(x^2+17x+60)(x^2+16x+60)-3x^2\)

\(=4(a+x)a-3x^2\) (đặt \(x^2+16x+60=a\))

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

\(=2a(2a-x)+3x(2a-x)=(2a-x)(2a+3x)\)

\(=(2x^2+32x+120-x)(2x^2+32x+120+3x)\)

\(=(2x^2+31x+120)(2x^2+35x+120)\)

\(=[2x(x+8)+15(x+8)](2x^2+35x+120)\)

\(=(2x+15)(x+8)(2x^2+35x+120)\)