a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)

\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)

\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)

\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)

\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)

\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)

b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)

\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)

\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)

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

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

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

Nếu bạn muốn có lời giải thì ít thôi @.@

Katherine Lilly Filbert nói rất đúng câu hỏi nhiều như vậy ai mà trả lời đc hết cơ chứ

x³+5x²+3x-9

=x³-x²+6x²-6x+9x-9

=x²(x-1)+6x(x-1)+9(x-1)

=(x-1)(x²+6x+9)

=(x-1)(x+3)²

x^3+5x^2+3x+9 = x^3+6x^2-x^2+9x-6x-9

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

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

Ta có

2x^4-x^3+2x^2+3x-2

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

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

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

Bạn ơi , tìm x 

câu b) )Đa thức <=> 
3.x^3 - x^2 - 6x^2 + 2x +15x - 5 
= x^2.(3x - 1) - 2x(3x-1) + 5(3x-1) 
= (3x-1).(x^2 - 2x +5) 

Tick nhé 

a> (x-2)(x^2+x+2)

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

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)\)

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

\(c,3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

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

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

\(\text{d) 2x}^4- 7x^3 - 2x^2 + 13x + 6\)
\(\text{= (2x^4 + 2x^3) - (9x^3 + 9x^2) + (7x^2 + 7x) + (6x + 6)}\)
\(\text{= 2x^3(x + 1) - 9x^2(x + 1) + 7x(x + 1) + 6(x + 1)}\)
\(\text{= (x + 1)(2x^3 - 9x^2 + 7x + 6)}\)
\(\text{= (x + 1)(2x + 1)(x - 3)(x - 2)}\)

phần b,c thay ''<=>'' là ''='' nhé ! Mình nhầm!

3x³-7x²+17x-5

=3x³-x²-6x²+2x+15x-5

= x².(3x-1)-2x.(3x-1)+5.(3x-1)

= (3x-1)(x²-2x+5)

Ta có : \(3x^3-7x^2+17x-5\)

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

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

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