\(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)

\(=\left(5x^4+4x^4\right)+\left(-3x^3+3x^3\right)+\left(x^2+x^2\right)-x+5\)

\(=9x^4+2x^2-x+5\)

\(Q\left(x\right)=x-5x_{}^3-x^2-x^4+4x^3-x^2+3x-1\)

\(=\left(-5x^4-x^4\right)+4x^3+\left(-x^2-x^2\right)+\left(x+3x\right)-1\)

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

P(x)+Q(x)

\(=9x^4+2x^2-x+5-6x^4+4x^3-2x^2+4x-1\)

\(=3x^4+4x^3+3x+4\)

Đặt P(x)+Q(x)=0

=>\(3x^4+4x^3+3x+4=0\)

=>\(x^3\left(3x+4\right)+\left(3x+4\right)=0\)

=>\(\left(3x+4\right)\left(x^3+1\right)=0\)

=>\(\left[\begin{array}{l}3x+4=0\\ x^3+1=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=-\frac43\\ x=-1\end{array}\right.\)

Thay giá trị x=2 vào đa thức B(x)=x²-4x+4, ta có:

\(2^2-4.2+4\)\(=4-8+4=0\)

Vậy x=2 là nghiệm của đa thức B(x)

Thay x = 2 vào đa thức B(x), ta có:

B(2) = 2² - 4.2 + 4

B(2) = 4 - 8 + 4

B(2) = - 4 + 4

B(2) = 0

Vậy x = 2 là nghiệm của đa thức B(x) = x² - 4x + 4.

\(x^5+x^4+x^3+x^2+x+1\)

\(=\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\)

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

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

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

Ta có: \(x^5+x^4+x^3+x^2+x+1\)

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

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

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

\(=\left(x+1\right)\left\lbrack\left(x^2+1\right)^2-x^2\right\rbrack=\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)