Phân tích đa thức thành nhân tử:
x4 - 5x2 + 4
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Ta có : \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
\(x^4+2x^3+2x^2+2x+1\\ =\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\\ =x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\\ =\left(x^3+x^2+x+1\right)\left(x+1\right)\\ =\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left(x^2+1\right)\left(x+1\right)^2\)
x 4 - 5 x 2 + 4 = x 4 - 4 x 2 - x 2 + 4 = x 4 - 4 x 2 - x 2 - 4 = x 2 x 2 - 4 - x 2 - 4 = x 2 - 4 x 2 - 1 = x + 2 x - 2 x + 1 x - 1
\(\text{x^3 – 5x^2 + 8x – 4 }\)
\(\text{= x^3 – 4x^2 + 4x – x^2 + 4x – 4}\)
\(\text{= x( x^2 – 4x + 4 ) – ( x^2 – 4x + 4 )}\)
\(\text{= ( x – 1 ) ( x – 2 )^2}\)
\(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x-4\\ =x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\\ =\left(x^2-4x+4\right)\left(x-1\right)\\ =\left(x-2\right)^2\left(x-1\right)\)
\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
\(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
`5x^2 + 5xy +x +y`
`=(5x^2 + 5xy )+(x+y)`
`=5x(x+y)+(x+y)`
`=(x+y)(5x+1)`
\(5x^2+5xy+x+y\\ =5x\left(x+y\right)+\left(x+y\right)\\ =\left(x+y\right)\left(5x+1\right)\)
45 + x 3 - 5 x 2 - 9 x = x 3 - 5 x 2 - 9 x - 45 = x 2 x - 5 - 9 x - 5 = x - 5 x 2 - 9 = x - 5 x - 3 x + 3
\(x^4-5x^2+4\)
\(=x^4-4x^2-x^2+4\)
\(=\left(x^4-4x^2\right)-\left(x^2-4\right)\)
\(=x^2\left(x^2-4\right)-\left(x^2-4\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)\)
\(x^4-5x^2+4\)
\(=x^4-4x^2-x+4\)
\(=x\left(x^3-1\right)-\left(4x^2-4\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)-4\left(x^2-1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left[x\left(x^2+x+1\right)-4\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x^3+x^2+x-4x+1\right)\)
\(=\left(x-1\right)\left(x^3+x^2-3x+1\right)\)
P/s : Có thể sai vì mk chưa soát lại bài , nên sai thông cảm !