\(=6x^2+9x+4x+6\)

\(=3x.\left(2x+3\right)+2.\left(2x+3\right)\)

\(=\left(2x+3\right).\left(3x+2\right)\)

\(\)

6x² + 13x + 6

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

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

6x²  +13x+6 = 6x²  + 4x +9x +6 = 2x(3x+2) + 3(3x+2) = (3x+2)(2x+3)

bạn bấm vào đúng 0 sẽ ra kết quả 

mình làm bài này rồi

Đừng tin bn Thạch bạn ấy nói dối đấy

2x^4 - 2x^3 +9x^3 - 9x^2 +7x^2 -  7x - 6x + 6 = 2x^3(x-1)+9x^2(x-1)+7x(x-1)-6(x-1) = (x-1)(2x^3+9x^2+7x-6)                                               =(x-1)(x+2)(x+3)(2x-1)

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

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

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

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

Tham Khảo

2x² + x - 6

= 2x² + 4x - 3x - 6

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

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

\(\left(x^2-2x-6\right)\left(x^2-2x-11\right)+6\)

\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+66+6\)

\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+72\)

\(=\left(x^2-2x-8\right)\left(x^2-2x-9\right)\)

\(=\left(x-4\right)\left(x+2\right)\left(x^2-2x-9\right)\)

Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)

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

\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)

\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)

6x² - 20x + 6

= 2(3x² - 10x + 3)

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

\(=6x^2-2x-18x+6\)

\(=2x.\left(3x-1\right)-6\left(3x-1\right)\)

\(=\left(2x-6\right).\left(3x-1\right)\)

=3x(x+2y)

=3x(x+2y)

3x² + 6xy

= 3x(x+2y)

x²-2x-15=(x²-5x)+(3x-15)=x(x-5)+3(x-5)=(x-5)(x+3)

\(x^2-3x-15=\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{69}{4}=\left(x-\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{69}}{2}\right)^2\)

\(=\left(x-\dfrac{3}{2}-\dfrac{\sqrt{69}}{2}\right)\left(x-\dfrac{3}{2}+\dfrac{\sqrt{69}}{2}\right)\)

\(x^2-2x-15=\left(x-5\right)\left(x+3\right)\)