Question

phân tích đa thức thành nhân tử

a,\(x^5+x+1\)

b,\(x^3+x^2+4\)

c\(x^4+2x^2-24\)

Answer 1

a. \(x^5+x+1\)

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

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

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

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

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

b.\(x^3+x^2+4\)

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

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

\(=\left(x+2\right)\left(x^2-x+2\right)\)
c.\(x^4+2x^2-24\)

\(=x^4+2x^3-2x^3-4x^2+6x^2+12x-12x-24\)

\(=x^3\left(x+2\right)-2x^2\left(x+2\right)+6x\left(x+2\right)-12\left(x+2\right)\)

\(=\left(x^3-2x^2+6x-12\right)\left(x+2\right)\)

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

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

Answer 2

a, x^5 + x + 1 = x ^ 5 - x^2 + (x ^2 + x + 1) = x^2 ( x-1) ( x^2+x+1) + ( x^2+x+1) = ( x^2+x+1 ) ( x^3-x^2+1)

c, x^4 + 2x^2 -24 = (x^4 +6x^2) - ( 4x^2+24) = x^2( x^2+6) - 4(x^2+6) = (x^2-4)(x^2 +6 ) = (x-2)(x+2)(x^2+6)