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

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

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

\(x^2-25=\left(x-5\right)\left(x+5\right)\)

\((x-1)(x-3)(x+8)\)

\(x^3+4x^2-29x+24\)

\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)

\(=\left(x+8\right)\left(x^2-4x+3\right)\)

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

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

a)  = x² - 3x + 2x - 6

    = x(x -3) + 2(x - 3)

    = (x - 3)(x + 2)

b)  = x² - x + 5x - 5

    = x(x - 1) + 5(x - 1)

    = (x - 1)(x + 5)

c)  = x³ - 5x² + 5x² - 25x + 6x - 30

    = x²(x - 5) + 5x(x - 5) +6(x - 5)

    = (x - 5)(x² + 5x + 6)

    = (x - 5)(x² + 2x + 3x + 6)

    = (x - 5)[x(x + 2) + 3(x + 2)]

    = (x - 5)(x + 2)(x + 3)

a)  = x² - 3x + 2x - 6

    = x(x -3) + 2(x - 3)

    = (x - 3)(x + 2)

b)  = x² - x + 5x - 5

    = x(x - 1) + 5(x - 1)

    = (x - 1)(x + 5)

c)  = x³ - 5x² + 5x² - 25x + 6x - 30

    = x²(x - 5) + 5x(x - 5) +6(x - 5)

    = (x - 5)(x² + 5x + 6)

    = (x - 5)(x² + 2x + 3x + 6)

    = (x - 5)[x(x + 2) + 3(x + 2)]

    = (x - 5)(x + 2)(x + 3)

\(a,x^4+4x^2-5\)

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

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

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

\(a^2+a+1=0\Rightarrow\left(a+\frac{1}{2}\right)^2+\frac{3}{4}=0\Rightarrow a\in C\)

Vì vậy P không tồn tại

Lớp 8 nên làm như này nhé :))

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

\(x^3-19x-30=x^3+6x-25x-30=x\left(x^2-25\right)+6x-30=x\left(x^2-25\right)+6\left(x-5\right)\)

\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)=\left(x-5\right)\left[\left(x\right)\left(x+5\right)+6\right]\)