1)  =\(-3x^4+9x^3+11x^3-33x^2-2x^2+6x-16x+48\)

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

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

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

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

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

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

Ý b lm theo ý tưởng tương tự nha bn :D 

\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)

\(=\left(5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x\right)+105\)

\(=5x\left(x^6+2x^5-4x^4-7x^3+4x^2-x+8\right)+105\)

Thay \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)vào đa thức ta được:

\(E=5x.0+105=105\)

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

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

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

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

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

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

Bài 1:

a: \(3x-6y=3\cdot x-3\cdot2y=3\left(x-2y\right)\)

b: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)

c: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=\left(2\cdot5x+2\cdot4y\right)\left(x-y\right)\)

\(=2\left(5x+4y\right)\left(x-y\right)\)

bài 2:

a: Đề thiếu vế phải rồi bạn

b: \(x^3-13x=0\)

=>\(x\left(x^2-13\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{13}\end{matrix}\right.\)

Bài 1:

a, $3x-6y$

$=3(x-2y)$

b, $14x^2y-21xy^2+28x^2y^2$

$=7xy(2x-3y+4xy)$

c, $10x(x-y)-8y(y-x)$

$=10x(x-y)-8y[-(x-y)]$

$=10x(x-y)+8y(x-y)$

$=(x-y)(10x+8y)$

$=2(x-y)(5x+4y)$

Bài 2:

a, Đề thiếu rồi bạn nhé.

b, \(x^3-13x=0\)

\(\Rightarrow x\left(x^2-13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\)