Question

[Đỗ Thị Hải Yến]

Phân tích thành nhân tử ( tách các hạng tử)

1,x^2-6x+3

2, 2m^2+10m+8

3, 9x^2+6x-8

4, x^3-5x^2-14x

5, a^4+4a^2-5

6, x^3-7x-6

7, 3x^2-22xy+4x+8y+7y^2+1

8, 12x^2+5x-12y^2+12y-10xy-3

Answer 1

[Kiệt Nguyễn]

1) \(x^2-6x+3\)

\(=x^2-6x+9-6\)

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

\(=\left(x-3+\sqrt{6}\right)\left(x-3-\sqrt{6}\right)\)

2) \(2m^2+10m+8\)

\(=2m^2+2m+8m+8\)

\(=2m\left(m+1\right)+8\left(m+1\right)\)

\(=\left(2m+8\right)\left(m+1\right)\)

\(=2\left(m+4\right)\left(m+1\right)\)

3) \(9x^2+6x-8\)

\(=\left(9x^2+6x+1\right)-9\)

\(=\left(3x+1\right)^2-9\)

\(=\left(3x+4\right)\left(3x-2\right)\)

4) \(x^3-5x^2-14x\)

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

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

\(=x\left[x\left(x-2\right)+7\left(x-2\right)\right]\)

\(=x\left(x+7\right)\left(x-2\right)\)

Answer 2

[lili]

1) x^2-6x+3

= (x^2-6x+9)-6

=(x-3)^2-6

=(x-3-căn 6)(x-3+căn 6)

2) =2(m^2+5m+4)

=2(m+1)(m+4)

3) =9x^2+6x+1-9

=(3x+1)^-9

=(3x-2)(3x+4)

4, x^3-5x^2-14x

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

5, a^4+4a^2-5

=a^4+4a^2+4-9

=(a^2+2)^-9

=(a^2-1)(a^2+5)

6, x^3-7x-6

=(x-3)(x+1)(x+2).