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

\(x^3+13x^2+x=x\left(x^2+13x+1\right)\)

\(=6x^2+3x+10x+5=3x\left(2x+1\right)+5\left(2x+1\right)=\left(3x+5\right)\left(2x+1\right)\)

6x2+3x+10x+5=3x(2x+1)+5(2x+1)=(3x+5)(2x+1)

ta có:

x^4+2014x^2+2013x+2014 = x^4+2013x^2+x^2+2013x+2013+1

                                        =(x^4+x^2+1)+2013(x^2+x+1)

                                       =(x^2+1)^2-x^2+2013(x^2+x+1)

                                       =(x^2-x+1)(x^2+x+1)+2013(x^2+x+1)

                                       =(x^2+x+1)(x^2+x+2014)

x⁴+2014x²+2013x+2014=(x⁴-x)+(2014x²+2014x+2014)

                                  =x(x-1)(x²+x+1)+2014(x²+x+1)

                                  =(x^2+x+1)(x²-x+2014)

\(P\left(x\right)=6x^3+13x^2+4x-3\)

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

\(=6x^2\left(x+1\right)+7x\left(x+1\right)-3\left(x+1\right)\)

\(=\left(6x^2+7x-3\right)\left(x+1\right)\)

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

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

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

\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)

\(3x^3-4x^2+13x-4\)

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

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

x³+13x²+7x-5

= x³+x²+12x²+12x-5x-5

=(x³+12x²-5x)+(x²+12x-5)

=x(x²+12x-5)+1(x²+12x-5)

=(x+1)(x²+12x-5)

phân tích đa thức thành nhân tử:

x³+13x²+7x-5

= x³+x²+12x²+12x-5x-5

=(x³+12x²-5x)+(x²+12x-5)

=x(x²+12x-5)+1(x²+12x-5)

=(x+1)(x²+12x-5)

Ai trên 10 điểm hỏi đáp thì mình nha mình đang cần gấp chỉ còn 99 điểm là tròn rồi mong các bạn hỗ trợ mình sẽ đền bù xứng đáng

tích nha 

tk cho nk

trần hoàng viêt nha , mk sẽ tk lại