Question

Tìm x thuộc Z để:

(4x³+11x²+5x+5)chia hết cho(x+2)

Answer 1

Ta có :

\(A=4x^3+11x^2+5x+5\)

\(=4x^3+8x^2+3x^2+6x-5x-10+15\)

\(=4x^2\left(x+2\right)+3x\left(x+2\right)-5\left(x+2\right)+15\)

\(=\left(x+2\right)\left(4x^2+3x-5\right)+15\)

\(\Leftrightarrow A\) chia hết cho \(x+2\) thì \(x+2\inƯ\left(15\right)\)

Ta có các TH :

+, \(x+2=1\Leftrightarrow x=-1\)

+, \(x+2=-1\Leftrightarrow x=-3\)

+, \(x+2=15\Leftrightarrow x=13\)

+, \(x+2=-15\Leftrightarrow x=-17\)

+, \(x+2=-3\Leftrightarrow x=-5\)

+, \(x+2=3\Leftrightarrow x=1\)

+, \(x+3=5\Leftrightarrow x=2\)

+, \(x+3=-5\Leftrightarrow x=-8\)

Vậy...