Question

Phân tích đa thức thành nhân tử

a) x^2-x-6

b) x^2-4x+3

Answer 1

\(\text{a) }x^2-x-6\\ =x^2-3x+2x-6\\ =\left(x^2-3x\right)+\left(2x-6\right)\\ =x\left(x-3\right)+2\left(x-3\right)\\ =\left(x-3\right)\left(x+2\right)\\ \\ \\ \)

\(\text{b) }x^2-4x+3\\ =x^2-x-3x+3\\ =\left(x^2-x\right)-\left(3x-3\right)\\ =x\left(x-1\right)-3\left(x-1\right)\\ =\left(x-3\right)\left(x-1\right)\\ \\ \)

Answer 2

a ) x² -x -6

= x² -x - 2²- 2

= (x-2).(x+2) -(x+2)

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

b) x² - 4x + 3 = x² - 4x + 4-1

= (x² - 2.2x +2² ) -1

= (x-2)² - 1

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

= ( x- 3).( x-1)