Question

 Cặp số (m;n) để đa thức Q(x)=m³ +(m-2)x² -(3n-5)x-4n đồng thời chia hết cho x+1 và x-3 là

Answer 1

Sửa đê: Q=mx^3+(m-2)x^2-(3n-5)x-4n

\(\dfrac{Q\left(x\right)}{x+1}\)

\(=\dfrac{mx^3+mx^2-2x^2-2x+\left(2-3n+5\right)x-4n}{x+1}\)
\(=mx^2-2x+\dfrac{\left(7-3n\right)x+7-3n-7-n}{x+1}\)

\(=mx^2-2x+7-3n+\dfrac{-n-7}{x+1}\)

Q(x) chia hết cho x+1

=>-n-7=0

=>n=-7

=>Q(x)=mx^3+(m-2)x^2+26x-28

\(\dfrac{Q\left(x\right)}{x-3}=\dfrac{mx^3-3mx^2+\left(4m-2\right)x^2-3\left(4m-2\right)x+\left(12m-6+26\right)x-28}{x-3}\)

\(=mx^2+\left(4m-2\right)x+\dfrac{\left(12m+20\right)x-28}{x-3}\)

\(=mx^2+\left(4m-2\right)x+\dfrac{\left(12m+20\right)x-3\left(12m+20\right)+3\left(12m+20\right)-28}{x-3}\)

\(=mx^2+\left(4m-2\right)x+12m+20+\dfrac{36m+32}{x-3}\)

Q(x) chia hết cho x-3

=>36m+32=0

=>m=-8/9