Question

Phân tích đa thức thành nhân tử

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

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

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

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

e) \(\left(x^2-8\right)^2+36\)

f) \(81x^4+4\)

Answer 1

a)

\((x^2+x)^2+3(x^2+x)+2\)

\(=(x^2+x)^2+(x^2+x)+2(x^2+x)+2\)

\(=(x^2+x)(x^2+x+1)+2(x^2+x+1)\)

\(=(x^2+x+2)(x^2+x+1)\)

b) \(x(x+1)(x+2)(x+3)+1\)

\(=[x(x+3)][(x+1)(x+2)]+1\)

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

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

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

Answer 2

c) \((x^2+x+1)(x^2+3x+1)+x^2\)

\(=(x^2+x+1)[(x^2+x+1)+2x]+x^2\)

\(=(x^2+x+1)^2+2x(x^2+x+1)+x^2\)

\(=(x^2+x+1+x)^2\)

\(=(x^2+2x+1)^2=[(x+1)^2]^2=(x+1)^4\)

d) \((x^2+1)^2-4x(1-x^2)\)

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

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

\(=(x^2+1)^2+(x-1)[4x^2+4+(4x-4)]\)

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

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

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

Answer 3

e) \((x^2-8)^2+36\)

\(=x^4-16x^2+100\)

\(=x^4+20x^2+100-36x^2\)

\(=(x^2+10)^2-(6x)^2\)

\(=(x^2+10-6x)(x^2+10+6x)\)

f) \(81x^4+4=(3x)^4+2^2=(9x^2)^2+2^2+2.(9x^2).2-2.(9x^2).2\)

\(=(9x^2+2)^2-4.9x^2=(9x^2+2)^2-(6x)^2\)

\(=(9x^2+2-6x)(9x^2+2+6x)\)