Question

tìm giá trị nguyên của x, để:

(2x^2+3x+2) chia hết cho (x+1)

Answer 1

Ta có : \(2x^2+3x+2=\left(2x^2+2x\right)+\left(x+1\right)+1\)

\(=2x\left(x+1\right)+\left(x+1\right)+1=\left(x+1\right)\left(2x+1\right)+1\)

Để \(\left(2x^2+3x+2\right)⋮\left(x+1\right)\)

thì \(1⋮x+1\) hay \(x+1\inƯ\left(1\right)\)

\(\Rightarrow x+1\in\left\{-1,1\right\}\)

\(\Leftrightarrow x\in\left\{-2,0\right\}\)

Vậy : \(x\in\left\{-2,0\right\}\) để \(\left(2x^2+3x+2\right)⋮\left(x+1\right)\)