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a) x2-xz-9y2+3yz
=(x2-9y2)-(xz-3yz)
=(x-3y)(x+3y)-z(x-3y)
=(x-3y)(x+3y-z)
b)x3-x2-5x+125
=x3-6x2+25x+5x2-30x+125
=x(x2-6x+25)+5(x2-6x+25)
=(x+5)(x2-6x+25)
c.x3+2x2-6x-27
=x3+5x2+9x-3x2-15x-27
=x(x2+5x+9)-3(x2+5x+9)
=(x-3)(x2+5x+9)
d. 12x3+4x2-27x-9
=12x3+4x2-27x-9
=4x2(3x+1)-9(3x+1)
=(4x2-9)(3x+1)
=(2x-3)(2x+3)(3x+1)
e.x4-25x2+20x-4
=x4+5x3-2x2-5x2-25x+10+2x2+10x-4
=x2(x2+5x-2)-5(x2+5x-2)+2(x2+5x-2)
=(x2-5x+2)(x2+5x-2)
f.x2(x2-6)-x2+9
=x4+x3-3x2-x3-x2+3x-3x2-3x+9
=x2(x2+x-3)-x(x2+x-3)-3(x2+x-3)
=(x2-x-3)(x2+x-3)
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\(25x^2-10x+1-16z^2=\left(5x-1-4z\right)\left(5x-1+4z\right)\)
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x4+4 = (x2)2+22 = x4 + 2.x2.2 + 4 – 4x2
= (x2 + 2)2 – (2x)2 = (x2-2x+2)(x2+2x+2)
Ta có: \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
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x4 + 4
= (x2)2 + 22
= x4 + 2.x2.2 + 4 – 4x2
(Thêm bớt 2.x2.2 để có HĐT (1))
= (x2 + 2)2 – (2x)2
(Xuất hiện HĐT (3))
= (x2 + 2 – 2x)(x2 + 2 + 2x)
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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
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\(x^4+x^3-20x^2-47x-15\)
\(=x^3\left(x-5\right)+6x^2\left(x-5\right)+10x\left(x-5\right)+3\left(x-5\right)\)
\(=\left(x-5\right)\left(x^3+6x^2+10x+3\right)\)
\(=\left(x-5\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)+\left(x+3\right)\right]\)
\(=\left(x-5\right)\left(x+3\right)\left(x^2+3x+1\right)\)
\(=x^4-5x^3+6x^3-30x^2+10x^2-50x+3x-15\\ =\left(x-5\right)\left(x^3+6x^2+10x+3\right)\\ =\left(x-5\right)\left(x^3+3x^2+3x^2+9x+x+3\right)\\ =\left(x-5\right)\left(x+3\right)\left(x^2+3x+1\right)\)