phân tích đa thức thành nhân tử
\(x^4+8x^2+7x+8\)
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a: \(x^4-6x^2+8\)
\(=x^4-4x^2-2x^2+8\)
\(=\left(x^2-2\right)\left(x^2-4\right)=\left(x^2-2\right)\left(x-2\right)\left(x+2\right)\)
b: \(x^4-5x^2-14\)
\(=x^4-7x^2+2x^2-14\)
\(=x^2\left(x^2-7\right)+2\left(x^2-7\right)=\left(x^2-7\right)\left(x^2+2\right)\)
c: \(4x^4-7x^2+3\)
\(=4x^4-4x^2-3x^2+3\)
\(=\left(x^2-1\right)\left(4x^2-3\right)=\left(x-1\right)\left(x+1\right)\left(4x^2-3\right)\)
d: \(6x^4+7x^2+2\)
\(=6x^4+3x^2+4x^2+2\)
\(=\left(2x^2+1\right)\left(3x^2+2\right)\)
e: \(x^4-8x^2+15\)
\(=x^4-5x^2-3x^2+15\)
\(=x^2\left(x^2-5\right)-3\left(x^2-5\right)=\left(x^2-5\right)\cdot\left(x^2-3\right)\)
a: \(x^4-2x^3+x^2-2x\)
\(=\left(x^4-2x^3\right)+\left(x^2-2x\right)\)
\(=x^3\left(x-2\right)+x\left(x-2\right)\)
\(=x\left(x-2\right)\left(x^2+1\right)\)
b: \(x^4+x^3-8x-8\)
\(=\left(x^4+x^3\right)-\left(8x+8\right)\)
\(=x^3\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-8\right)\)
\(=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)
a) 3x2 - 7x + 4
= 3x2 - 3x - 4x + 4
= 3x( x - 1 ) - 4( x - 1 )
= ( x - 1 )( 3x - 4 )
b) x2 - 6xy + 9y2 = ( x - 3y )2
c) x2 - 8x - 9
= x2 - 9x + x - 9
= x( x - 9 ) + ( x - 9 )
= ( x - 9 )( x + 1 )
a) 3x2 - 7x + 4
= 3x2 - 4x - 3x + 4
= (3x2 - 4x) - (3x - 4)
= x.(3x - 4) - (3x - 4)
= (3x - 4).(x - 1)
b) x2 - 6xy + 9y2
= x2 - 2.x.3y + (3y)2
= (x - 3y)2
c) x2 - 8x - 9
= x2 - 9x + x - 9
= (x2 - 9x) + (x - 9)
= x.(x - 9) + (x - 9)
= (x - 9).(x + 1)
Ta có : x4 + 8x2 + 7x + 8
= x4 - x + 8x2 + 8x + 8
= x(x3 - 1) + 8(x2 + x + 1)
= x(x - 1)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 - x)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 8)
Học tốt nhé !