Question

Phân tích đa thức sau thành nhân tử

a,(x-1)(x+2)(x+3)

b,(x²-11x+28)(x²-7x+10)-72

c,4x³+x²+x-3

Answer 1

b,(x²-11x+28)(x²-7x+10)-72

(x-4)(x-7)(x-2)(x-5)-72

(x-2)(x-7)(x-4)(x-5)-72

(x²-9x+14)(x²-9x+20)-72

Đặt"t"=x²-9x+14, ta có:

(=)t(t+6)-72

(=)t²+6t-72

(=)t²+12t-6t-72

(=)t(t-6)-12(t-6)

(=)(t-6)(t-12)

(=)(x²-9x+14-6)(x²-9x+14-12)

(=)(x²-9x+8)(x²-9x+2)

Answer 2

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

b: \(=\left(x-4\right)\left(x-7\right)\left(x-2\right)\left(x-5\right)-72\)

\(=\left(x^2-9x+20\right)\left(x^2-9x+14\right)-72\)

\(=\left(x^2-9x\right)^2+34\left(x^2-9x\right)+208\)

\(=\left(x^2-9x+26\right)\left(x^2-9x+8\right)\)

\(=\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)