Question

tìm a để (6x^3-x^2-ax-2) chia hết (2x-3)

e tính ra 3x^2+4x dư 12x-ax-2 r ak

Answer 1

\(\dfrac{6x^3-x^2-ax-2}{2x-3}\)

\(=\dfrac{6x^3-9x^2+8x^2-12x+\left(12-a\right)x-1,5\left(12-a\right)+1,5\left(12-a\right)-2}{2x-3}\)

\(=\dfrac{3x^2\left(2x-3\right)+4x\left(2x-3\right)+\left(12-a\right)\left(x-1,5\right)+18-1,5a-2}{2x-3}\)

\(=3x^2+4x+\dfrac{1}{2}\left(12-a\right)+\dfrac{-1,5a+16}{2x-3}\)

Để \(6x^3-x^2-ax-2⋮2x-3\) thì -1,5a+16=0

=>-1,5a=-16

=>\(a=\dfrac{16}{1,5}=\dfrac{32}{3}\)