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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)
\(4x^2-7x-2\\ =4x^2-8x+x-2\\ =4x\left(x-2\right)+\left(x-2\right)\\ =\left(x-2\right)\left(4x+1\right)\)
a: \(=x\left(x^2+4x+4-z^2\right)\)
\(=x\left(x+2-z\right)\left(x+2+z\right)\)
Ta có:
\(\left(x^4+2x^3-x-2\right)+\left(4x^2+4x+4\right)\)
\(=\left[\left(x^4+2x^3\right)-\left(x+2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left[x^3\left(x+2\right)-\left(x-2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+1\right)+4\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x-1\right)\left(x+2\right)+4\right]\)
\(=\left(x^2+x+1\right)\left(x^2+x+2\right)\)
x3 – 4x2 – 12x + 27
(Nhóm để xuất hiện nhân tử chung)
= (x3 + 27) – (4x2 + 12x)
= (x3 + 33) – (4x2 + 12x)
(nhóm 1 là HĐT, nhóm 2 có 4x là nhân tử chung)
= (x + 3)(x2 – 3x + 9) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9 – 4x)
= (x + 3)(x2 – 7x + 9)
Ta thấy:
\(12x^3+4x^2-27x-9\)
\(=12x^3-18x^2+22x^2-33x+6x-9\)
\(=6x^2\left(2x-3\right)+11x\left(2x-3\right)+3\left(2x-3\right)\)
\(=\left(6x^2+11x+3\right)\left(2x-3\right)\)
\(=\left(6x^2+9x+2x+3\right)\left(2x-3\right)\)
\(=\left[3x\left(2x+3\right)+\left(2x+3\right)\right]\left(2x-3\right)\)
\(=\left(3x+1\right)\left(2x+3\right)\left(2x-3\right)\)