Question

Phân tích thành nhân tử theo phương pháp nhóm hạng tử :

a) x³ - x² - 5x + 125

b) x³ + 2x² - 6x - 27

c) 12x³ + 4x² - 27x - 9

Answer 1

a) x³ - x² - 5x + 125

=(x³-6x²+25x)+(5x²-30x+125)

=x(x²-6x+25)+5(x²-6x+25)

=(x+5)(x²-6x+25)

b) x³ + 2x² - 6x - 27

=x³+5x²+9-3x²-15x-27

=x(x²+5x+9)-3(x²+5x+9)

=(x-3)(x²+5x+9)

c) 12x³ + 4x² - 27x - 9

=4x²(3x+1)-9(3x+1)

=(4x²-9)(3x+1)

=[(2x)²-3²](3x+1)

=(2x-3)(2x+3)(3x+1)

 

Answer 2

a) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

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

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

b) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)

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

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

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

c) \(12x^3+4x^2-27x-9\)

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

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

Answer 3

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

\(=\left(x+5\right)\left(x^3-5x+25\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^3-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^3-6x+25\right)\)

Answer 4

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

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

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

\(=\left(5+x\right)\left(25-4x+x^2\right)\)

Answer 5

b) \(x^3+2x^2-6x-27\)

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

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

\(=\left(x+3\right)\left(x^3+5x+9\right)\)

Answer 6

a) x³ - x² - 5x + 125

=\(x^3+5x^2-6x^2-30x+15x+125\)

=\(\left(x+5\right)\left(x^2-6x+15\right)\)

b) x³ + 2x² - 6x - 27

=\(x^3-3x^2+5x^2-15x-9-27\)

=\(\left(x-3\right)\left(x^2+5x-9\right)\)

c) 12x³ + 4x² - 27x - 9

=\(\left(x-\frac{3}{2}\right)\left(x+\frac{1}{3}\right)\left(x+\frac{3}{2}\right)\)

=\(\)

Answer 7

c) \(12x^3+4x^2-27x-9\)

\(=12x^3+\left(2x\right)^2-3^3x-3^2\)

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

\(=\left(2x-3\right)\left(2x+3\right)+3\left\{x\left[\left(2x\right)^2-3^2\right]\right\}\)

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

\(\left(2x-3\right)\left(2x+3\right)\left(3x+1\right)\)