Question

x^2×(x+1)^2+2x^2+2x—8 phân tích đa thưc thành nt

Answer 1

x²( x + 1 )² + 2x² + 2x - 8

= [ x( x + 1 ) ]² + 2( x² + x ) - 8

= ( x² + x )² + 2( x² + x ) - 8 (*)

Đặt a = x² + x

(*) = a² + 2a - 8

= a² - 2a + 4a - 8

= a( a - 2 ) + 4( a - 2 ) 

= ( a - 2 )( a + 4 )

= ( x² + x - 2 )( x² + x + 4 )

= ( x² - x + 2x - 2 )( x² + x + 4 )

= [ x( x - 1 ) + 2( x - 1 ) ]( x² + x + 4 )

= ( x - 1 )( x + 2 )( x² + x + 4 )

Answer 2

ta co

\(x^2\left(x+1\right)^2+2x^2+2x-8\)

=\(\left(x\left(x+1\right)\right)^2+2x\left(x+1\right)+1-9\)

=\(\left(x^2+x+1\right)^2-9\)

=\(\left(x^2+x-8\right)\left(x^2+x+10\right)\)

=\(\left(x^2+2x\frac{1}{2}+\frac{1}{4}-\frac{33}{4}\right)\left(x^2+x+10\right)\)

=\(\left(\left(x+\frac{1}{2}\right)^2-\frac{33}{4}\right)\left(x^2+x+10\right)\)

=\(\left(x+\frac{1}{2}-\sqrt{\frac{33}{4}}\right)\left(x+\frac{1}{2}+\sqrt{\frac{33}{4}}\right)\left(x^2+x+10\right)\)

Answer 3

mik xin lỗi

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

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

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