Question

Phân tích các đa thức sau thành nhân tử:

a)x^2 + 9x + 20 b)x^4 - 5x^2 + 4 c)x^4 + 4 d)x(x + 1)(x + 2)(x + 3) + 1

Answer 1

\(a,x^2+9x+20=x^2+4x+5x+20.\)

\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)

\(b,x^4-5x^2+4=x^4-x^2-4x^2+4\)

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

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

\(c,x^4+4=x^4+4x^2+4-4x^2\)

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

\(d,x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

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

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

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

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

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