Phân tích đa thức sau thành nhân tử: \(x^8+3x^4+\text{4}\)
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a: \(x^4+x^2+2x+6\)
\(=x^4-2x^3+3x^2+2x^3-4x^2+6x+2x^2-4x+6\)
\(=\left(x^2-2x+3\right)\left(x^2+2x+2\right)\)
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\(x^4-3x^3-x+3\)
\(=x^4-x^3-2x^3+2x-3x+3\)
\(=\)\(x^3\left(x-1\right)-2x\left(x^2-1\right)-3\left(x-1\right)\)
\(=x^3\left(x-1\right)-2x\left(x-1\right)\left(x+1\right)-3\left(x-1\right)\)
\(=\left[x^3-2x\left(x+1\right)-3\right]\left(x-1\right)\)
\(=\left[x^3-2x^2-2x-3\right]\left(x-1\right)\)
\(=\)\(\left[x^3-3x^2+x^2-3x+x-3\right]\left(x-1\right)\)
\(=\left[x^2\left(x-3\right)+x\left(x-3\right)+\left(x-3\right)\right]\left(x-1\right)\)
\(=\left[\left(x-3\right)\left(x^2+x+1\right)\right]\left(x-1\right)\)
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a) x8 + x + 1 = (x^2+x+1)*(x^6-x^5+x^3-x^2+1)
b) x^8 + 3x^4 + 4 = (x^4-x^2+2)*(x^4+x^2+2)
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\(x^3-3x^2+6x-4\)
\(=x^3-2x^2+4x-x^2+2x-4\)
\(=\left(x^3-2x^2+4x\right)-\left(x^2-2x+4\right)\)
\(=x\left(x^2-2x+4\right)-\left(x^2-2x+4\right)\)
\(=\left(x-1\right)\left(x^2-2x+4\right)\)
x^3 - 3x^2 + 6x - 4
<=> x^3-3x^2+3x-1+3x-3
<=>(x-1)^3+3(x-1)
<=>(x-1)+((x-1)^2+3)
<=>(x-1)+(x^2-2x+4)
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\(x^8+3x^4+4\)
\(=x^8+4x^4+4-x^4\)
\(=\left(x^4-2\right)^2-x^4\)
\(=\left(x^4-x^2-2\right)\left(x^4-x^2-2x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+1\right)\left(x^2-1\right)\left(x^2+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-2\right)\left(x^2+1\right)\left(x^2+2\right)\)
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Trả lời:
a, x4 + 3x3 + x2 + 3x
= ( x4 + 3x3 ) + ( x2 + 3x )
= x3 ( x + 3 ) + x ( x + 3 )
= ( x3 + x ) ( x + 3 )
= x ( x2 + 1 ) ( x + 3 )
b, Sửa đề: x4 - x2 + 8x - 8
= ( x4 - x2 ) + ( 8x - 8 )
= x2 ( x2 - 1 ) + 8 ( x - 1 )
= x2 ( x - 1 ) ( x + 1 ) + 8 ( x - 1 )
= ( x - 1 ) [ x2 ( x + 1 ) + 8 ]
= ( x - 1 ) ( x3 + x2 + 8 )
x8 + 3x4 + 4
= ( x8 + 4x4 + 4 ) - x4
= ( x4 + 2 )2 - ( x2 )2
= ( x4 - x2 + 2 )( x4 + x2 + 2 )
\(=x^8+4x^4+4-x^4\)
\(=\left(x^4-2\right)^2-x^4\)
\(=\left(x^4-x^2-2\right)\left(x^4+x^2-2\right)\)
\(=\left(x^4-2x^2+x^2-2\right)\left(x^4+2x^2-x^2-2\right)\)
\(=\left(x^2\left(x^2-2\right)+x^2-2\right)\left(x^2\left(x^2+2\right)-\left(x^2+2\right)\right)\)
\(=\left(x^2+1\right)\left(x^2-2\right)\left(x^2-1\right)\left(x^2+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^2-2\right)\left(x^2+2\right)\)