Question

tìm a

a,4x^2+3x+a chia hết cho x+3

Answer 1

\(4x^2+3x+a=4x\left(x+3\right)-9\left(x+3\right)+27+a=\left(x+3\right)\left(4x-9\right)+27+a⋮x+3\)

\(\Rightarrow27+a=0\Rightarrow a=-27\)

Answer 2

\(4x^2+3x+a⋮x+3\\ \Leftrightarrow4x^2+3x+a=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow4\cdot9-9+a=0\\ \Leftrightarrow27+a=0\\ \Leftrightarrow a=-27\)