a) x3 - 4x2 - 12x + 27
= \(\left(x^3+3x^2\right)-\left(7x^2+21x\right)+\left(9x+27\right)\)
= \(\left(x+3\right)\left(x^2-7x+9\right)\)
b) 9x2 + 6x - 8
=\(9x^2-6x+12x-8=3x\left(3x-2\right)+4\left(3x-2\right)\)
=\(\left(3x-2\right)\left(3x+4\right)\)
c) x2 - 7xy + 10y2
=\(x^2-5xy-2xy+10y^2=x\left(x-5y\right)-2y\left(x-5y\right)\)
=\(\left(x-5y\right)\left(x-2y\right)\)
a) x3 - 4x2 - 12x + 27
=x3 + 3x2 - 7x2 - 21x + 9x + 27
= x2(x+3) - 7x(x+3) + 9(x+3)
= (x2 - 7x + 9)(x + 3)
b) 9x2 + 6x - 8
= 9x2 - 6x + 12x - 8
= 3x(3x - 2) + 4(3x - 2)
= (3x + 4)(3x - 2)
c) x2 - 7xy + 10y2
= x2 - 5xy - 2xy + 10y2
= x(x - 5y) - 2y(x - 5y)
= (x - 2y)(x - 5y)
d) x8 + x7 + 1
Ta thêm vào các số hạng x6, x5, x4, x3, x2, x và cùng bớt đi các số hạng ấy ta có:
= x8 - x6 + x5 - x3 + x2 + x7 - x5 + x4 -x2 +x + x6 - x4 + x3 - x + 1
= x2(x6 - x4 + x3 - x + 1) + x(x6 - x4 + x3 - x + 1) + x6 - x4 + x3 - x + 1
= (x2 + x + 1)(x6 - x4 + x3 - x + 1)
a) \(x^3-4x^2-12x+27\)
\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
b) \(9x^2+6x-8\)
\(=9x^2+6x+1-9\)
\(=\left(3x+1\right)^2-9\)
\(=\left(3x+1-9\right)\left(3x+1+9\right)\)
\(=\left(3x+8\right)\left(3x+10\right)\)
c) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-5\right)\)