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\(x^3+x^2+9x-10x^2-10x+25x+25\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+25\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+25\right)=\left(x+1\right)\left(x-5\right)^2\)
a) x3 - 6x2 + 11x - 6
= ( x3 - 2x2 ) - ( 4x2 - 8x ) + ( 3x - 6 )
= x2( x - 2 ) - 4x( x - 2 ) + 3( x - 2 )
= ( x - 2 )( x2 - 4x + 3 )
= ( x - 2 )( x2 - x - 3x + 3 )
= ( x - 2 )[ x( x - 1 ) - 3( x - 1 ) ]
= ( x - 2 )( x - 1 )( x - 3 )
b) x3 - 6x2 - 9x + 14
= ( x3 - x2 ) - ( 5x2 - 5x ) - ( 14x - 14 )
= x2( x - 1 ) - 5x( x - 1 ) - 14( x - 1 )
= ( x - 1 )( x2 - 5x - 14 )
= ( x - 1 )( x2 + 2x - 7x - 14 )
= ( x - 1 )[ x( x + 2 ) - 7( x + 2 ) ]
= ( x - 1 )( x + 2 )( x - 7 )
c) x3 + 6x2 + 11x + 6
= ( x3 + 2x2 ) + ( 4x2 + 8x ) + ( 3x + 6 )
= x2( x + 2 ) + 4x( x + 2 ) + 3( x + 2 )
= ( x + 2 )( x2 + 4x + 3 )
= ( x + 2 )( x2 + x + 3x + 3 )
= ( x + 2 )[ x( x + 1 ) + 3( x + 1 ) ]
= ( x + 2 )( x + 1 )( x + 3 )
e) x6 - 9x3 + 8
Đặt t = x3
bthuc <=> t2 - 9t + 8
= t2 - t - 8t + 8
= t( t - 1 ) - 8( t - 1 )
= ( t - 1 )( t - 8 )
= ( x3 - 1 )( x3 - 8 )
= ( x - 1 )( x2 + x + 1 )( x - 2 )( x2 + 2x + 4 )
\(x^3+6x^2+11x+6=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
6x3 - 11x2 - x - 2
= 6x3 - 12x2 + x2 - 2x + x - 2
= ( 6x3 - 12x2 ) + ( x2 - 2x ) + ( x - 2 )
= 6x2( x - 2 ) + x( x - 2 ) + 1( x - 2 )
= ( x - 2 )( 6x2 + x + 1 )
x3+6x2+11x+6 = x3+6x2+12x-x+8-2 = (x3+6x2+12x+8) - (x+2) = (x+2)3 - (x+2) = (x+2)[(x+2)2 - 1] = (x+2)(x+2-1)(x+2+1) = (x+2)(x+1)(x+3)
Ta có:\(x^3+9x^2+11x-21\)
\(=x^3-x^2+10x^2-10x+21x-21=x^2\left(x-1\right)+10x\left(x-1\right)+21\left(x-1\right)\)
\(=\left(x^2+10x+21\right)\left(x-1\right)=\left(x^2+3x+7x+21\right)\left(x-1\right)\)
\(=\left[x\left(x+3\right)+7\left(x+3\right)\right]\left(x-1\right)\)
\(=\left(x+3\right)\left(x+7\right)\left(x-1\right)\)
x^3+9x^2+11x-21=x^3-x^2+10x^2-10x+21x-21=(x^3-x^2)+(10x^2-10x)+(21x-21)
=x^2(x-1)+10x(x-1)+21(x-1)=(x-1)(x^2+10x+21)=(x-1)(x^2+3x+7x+21)=(x-1)[(x^2+3x)+(7x+21)]
=(x-1)(x+7)(x+3)