phân tích đa thức thành nhân tử dạng đoán nghiêmk 1, x^3+3x^2-10x-24 2, 2x^3-11x^2+10x+8 3, 2x^3+11x^2+3x-36 4, 6x^3-17x^2-4x+3

\(x^3+3x^2-10x-24\)

\(=x^3-3x^2+6x^2-18x+8x-24\)

\(=x^2\left(x-3\right)+6x\left(x-3\right)-8\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+6x-8\right)\)

\(=\left(x-3\right)\left(x^2+6x+9-1\right)\)

\(=\left(x-3\right)[\left(x-3\right)^2-1]\)

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

\(2x^3-11x^2+10x+8\)

\(=2x^3-4x^2-7x^2+14x-4x+8\)

\(=2x^2\left(x-2\right)-7x\left(x-2\right)-4\left(x-2\right)\)

\(=\left(x-2\right)\left(2x^2-7x-4\right)\)

\(=\left(x-2\right)[2x\left(x-4\right)+\left(x-4\right)]\)

\(=\left(x-2\right)\left(x-4\right)\left(2x+1\right)\)

Ta có: \(^{3x^3-4x^2+13x-4}\) = \(3x^3-x^2-3x^2+x+12x-4\)

                                       = \(3x^2\left(x-\frac{1}{3}\right)-3x\left(x-\frac{1}{3}\right)+12\left(x-\frac{1}{3}\right)\)

                                       = \(\left(3x^2-3x+12\right)\left(x-\frac{1}{3}\right)\)

                                       = \(3\left(x^2-x+4\right)\left(x-\frac{1}{3}\right)\)

 3x^3-4x^2+13x-4

= (3x-1)(x^2-x+4)

nha bạn 

\(2x^3-35x+75=2x^3+10x^2-10x^2-50x+15x+75\)

\(=\left(x+5\right)\left[2x^2-10x+15\right]\)

Trả lời:

Sửa đề: \(2x^2-35x+75\)

\(=2x^2-30x-5x+75\)

\(=\left(2x^2-30x\right)-\left(5x-75\right)\)

\(=2x\left(x-15\right)-5\left(x-15\right)\)

\(=\left(x-15\right)\left(2x-5\right)\)

a. Đặt \(x^2-2y=a\)

ta có : \(\left(x^2-2y\right)^2-4\left(x^2-2y\right)-12=a^2-4a-12=a^2-6a+2a-12=\left(a-6\right)\left(a+2\right)\)

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

b. Đặt \(x+6=a\Rightarrow\left(x+3\right)\left(x+6\right)\left(x+9\right)+45=\left(a-3\right)a\left(a+3\right)+45\)

\(=a^3-9a+45\) nghiệm xấu quá không nhóm được ban ơi :((

Giúp em với ạ huhu

Ta có (x³ + ax² + bx + 4) : (x - 2)² = x + a + 4 dư (4a + b + 12)x - 4(a + 3) 

(x³ + ax² + bx + 4) \(⋮\) (x - 2)² khi \(\hept{\begin{cases}4a+b+12=0\\a+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}b=0\\a=-3\end{cases}}\)

Vậy b = 0 ; a = -3 

\(1,2x^3+3x^2-8x+3\)

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

\(=2x^2\left(x-1\right)+5x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(2x^2+5x-3\right)\left(x-1\right)\)

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

\(2,x^3-5x^2+2x+8\)

\(=x^3+x^2-6x^2-6x+8x+8\)

\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)

\(=\left(x^2-6x+8\right)\left(x+1\right)\)

\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\)

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

\(=-6x^3-6x^2+7x^2+7x-2x-2\)

\(=-6x^2\left(x+1\right)+7x\left(x+1\right)-2\left(x+1\right)\)

\(=\left(-6x^2+7x-2\right)\left(x+1\right)\)

\(=\left(-6x^2-3x-4x-2\right)\left(x+1\right)\)

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

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

\(4,3x^3+19x^2+4x-12\)

\(=3x^3+18x^2+x^2+6x-2x-12\)

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

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

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