3x^2+mx+27 x+5 3x+(m-15) 3x^2+15x - (m-15)x+27 (m-15)x+5(m-15) - 27-5(m-15)

Vì \(A\left(x\right):B\left(x\right)\)dư 2 \(\Leftrightarrow27-5\left(m-15\right)=2\)

                                           \(\Leftrightarrow m-15=5\)

                                           \(\Leftrightarrow m=20\)

Vậy ...

Câu 4: 

Để f(x) chia hết cho g(x) thì \(x^2+5x+a⋮x+1\)

\(\Leftrightarrow x^2+x+4x+4+a-4⋮x+1\)

=>a-4=0

hay a=4

Câu 5: 

Đêt f(x) chia hết cho g(x) thì \(2x^2+3x+a⋮x+2\)

\(\Leftrightarrow2x^2+4x-x-2+a+2⋮x+2\)

=>a+2=0

hay a=-2

A(x) chia cho B(x) có số dư bằng 2 nên 102 – 5m = 2 ⇒ -5m = 100

⇒ m = 20

help me! 

Tìm m để

a, (x^4+5x^3-x^2-17x+m+4)chia hết cho (x^2+2x-3)

b, (2x^4+mx^3-mx-2) chia hết cho (x^2-1)

1: \(3x^2+mx+n\)

\(=3x^2+15x+\left(m-15\right)x+5m-75-5m+75+n\)

\(3x^2+mx+n\) chia x+5 dư 27

=>-5m+75+n=27

=>-5m+n=27-75=-48

2: \(x^3+mx+n\)

\(=x^3+x^2-x^2-x+\left(m+1\right)x+m+1+n-m-1\)

\(=\left(x+1\right)\left(x^2-x+m+1\right)+n-m-1\)

\(x^3+mx+n\) chia x+1 dư 7

=>n-m-1=7

=>n=m+1+7=m+8

\(x^3+mx+n\)

\(=x^3-3x^2+3x^2-9x+\left(m+9\right)x+3m+27+n-3m-27\)

\(=\left(x-3\right)\left(x^2+3x+m+9\right)+n-3m-27\)

\(x^3+mx+n\) chia x-3 dư 5

=>n-3m-27=5

=>n=3m+27+5=3m+32

mà n=m+8

nên 3m+32=m+8

=>2m=-24

=>m=-12

n=m+8=-12+8=-4

\(f\left(x\right)⋮g\left(x\right)\Leftrightarrow3x^2-mx-2=\left(x-m\right)\cdot a\left(x\right)\)

Thay \(x=m\)

\(\Leftrightarrow3m^2-m^2-2=0\\ \Leftrightarrow2m^2=2\Leftrightarrow m^2=1\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-1\end{matrix}\right.\)