Question

Phân tích đa thức thành nhân tử

a, $(x^2+x)^2 -5(x^2+x)+6$

b, $(x^2+2x)^2-2(x+1)^2-1$

Answer 1

\(a,\left(x^2+x\right)^2-5\left(x^2+x\right)+6\\ =\left[\left(x^2+x\right)^2-2\left(x^2+x\right)\right]+\left[-3\left(x^2+x\right)+6\right]\\ =\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\\ =\left(x^2+x-2\right)\left(x^2+x-3\right)\\ b,\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\\ =\left(x^2+2x\right)^2-2\left(x^2+2x+1\right)-1\\ =\left(x^2+2x\right)^2-2\left(x^2+2x\right)-2-1\\ =\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3\\ =\left[\left(x^2+2x\right)^2-3\left(x^2+2x\right)\right]+\left[\left(x^2+2x\right)-3\right]\\ =\left(x^2+2x\right)\left(x^2+2x-3\right)+\left(x^2+2x-3\right)\\ =\left(x^2+2x-3\right)\left(x^2+2x+1\right)\\ =\left(x-1\right)\left(x+3\right)\left(x+1\right)^2\)

Answer 2

a) \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

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

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

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

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

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

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

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

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

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

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

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

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

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