Bài 8: Phân tích đa thức thành nhân tử bằng phương pháp nhóm các hạng tử

ND

Phân tích đa thức thành nhân tử

a) \(4x^4+y^4\)

b) \(\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27\)

c) \(x^3-x^2-5x+125\)

d) \(xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz\)

TH
31 tháng 8 2018 lúc 20:55

a) \(4x^4+y^4\)

\(=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-2.2x^2.y^2\)

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

\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)

\(=\left(2x^2+y^2+2xy\right)\left(2x^2+y^2-2xy\right)\)

b) \(\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27\)

\(=\left(x^2-3x-1\right)^2-2\left(x^2-3x-1\right).6+36-9\)

\(=\left(x^2-3x-1-6\right)^2-3^2\)

\(=\left(x^2-3x-7\right)^2-3^2\)

\(=\left(x^2-3x-7-3\right)\left(x^2-3x-7+3\right)\)

\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

c) \(x^3-x^2-5x+125\)

\(=x^3+5x^2-6x^2-30x+25x+125\)

\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

d) \(xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz\)

\(=xy\left(x+y\right)+yz\left(y+z\right)+xyz+zx\left(z+x\right)+xyz\)

\(=xy\left(x+y\right)+yz\left(y+z+x\right)+zx\left(z+x+y\right)\)

\(=xy\left(x+y\right)+z\left(x+y+z\right)\left(y+x\right)\)

\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)

\(=\left(x+y\right)\left(xy+zx+yz+z^2\right)\)

\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)

\(=\left(x+y\right)\left(x+z\right)\left(y+z\right)\)

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MP
31 tháng 8 2018 lúc 21:07

a) ta có : \(4x^4+y^4=4x^4+4x^2y^2+y^2-\left(2xy\right)^2\)

\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)

b) ta có : \(\left(x^3-3x-1\right)^2-12\left(x^2-3x-1\right)+27\)

\(=\left(x^2-3x-1\right)^2-3\left(x^2-3x-1\right)-9\left(x^2-3x-1\right)+27\)

\(=\left(x^2-3x-1\right)\left(x^2-3x-4\right)-9\left(x^2-3x-4\right)\)

\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

c) ta có : \(x^3-x^2-5x+125=x^2+5x^2-6x^2-30x+25x+125\)

\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)=\left(x^2-6x+25\right)\left(x+5\right)\)

d) ta có : \(xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz\)

\(=x^2y+xy^2+y^2z+xyz+yz^2+z^2x+zx^2+xyz\)

\(=y\left(x^2+xy+yz+xz\right)+z\left(yz+zx+x^2+xy\right)\)

\(=\left(x+y\right)\left(x^2+xy+yz+xz\right)\)

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