Question

Cho C = \(3x^3+2x^2-4x+a+49\)

D = \(x+3\)

Tìm a để C chia hết cho D

Answer 1

Để C chia hết cho D thì \(3x^3+9x^2-7x^2-21x+17x+51+a-2⋮x+3\)

=>a-2=0

=>a=2

Answer 2

\(C=3x^3+2x^2-4x+a+49\)

\(=\left(3x^3+9x^2\right)-\left(7x^2+21x\right)+\left(17x+51\right)+a-2\)

\(=3x^2\left(x+3\right)-7x\left(x+3\right)+17\left(x+3\right)+a-2\)

\(=\left(3x^2-7x+17\right)\left(x+3\right)+a-2\)

Để \(C⋮D=x+3\)

\(\Leftrightarrow a-2=0\Leftrightarrow a=2\)