(2\(x^3\) - 3\(x^2\) + \(x\) + a) ⋮ (\(x\) + 2)
Theo bezout ta có:
(2\(x^3\) - 3\(x^2\) + \(x\) + a) ⋮ (\(x+2\)) ⇔
F(\(x\)) = (2\(x^3\) - 3\(x^2\) + \(x\) + a) thỏa mãn F(-2) = 0
Thay \(x\) = - 2 vào F(\(x\)) ta có:
2.(-2)3 - 3.22 + (-2) + a = 0
2.(-8) - 3.4 - 2 + a = 0
a = 2.8 + 3.4 + 2
a = 16 + 12 + 2
a = 28 + 2
a = 30
\(2x^3-3x^2+x+a=2x^2-7x+15+\dfrac{a-30}{x+2}\)
Để \(2x^3-3x^2+x+a⋮x+2\)
\(\Leftrightarrow a-30=0\)
\(\Leftrightarrow a=30\)