Question

[Trần Mỹ Chi]

Help me ~~~

Phân tích đa thức thành nhân tử

a)x^4-3x^3+4x^2-3x+1

b)x^4+2021x^2-2020x+2021

c)6x^4+5x^3-38x^2+5x+6

Nhanh mk tick

Answer 1

[FL.Hermit]

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

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

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

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

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

c)

\(=6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6\)

\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)

\(=\left(6x^3+18x^2-x^2-3x-x-3\right)\left(x-2\right)\)

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

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

Answer 2

[FL.Hermit]

b)

\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)

\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)

\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)

CÓ:   \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)

=>   \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)

=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.