1: \(\dfrac{f\left(x\right)}{x-3}=\dfrac{2x^2-6x+\left(a+6\right)x-3a-18+3a+19}{x-3}\)

=2x^2+(a+6)+3a+19/x-3

Để f(x)/x-3 dư 4 thì 3a+19=4

=>3a=-15

=>a=-5

2: \(\dfrac{f\left(x\right)}{x-5}=\dfrac{3x^2-15x+\left(a+15\right)x-5a-75+5a+102}{x-5}\)

\(=3x+a+15+\dfrac{5a+102}{x-5}\)

Để dư là 27 thì 5a+102=27

=>5a=-75

=>a=-15

a: \(\Leftrightarrow2x^2+8x+\left(a-8\right)x+4\left(a-8\right)-4a+28⋮x+4\)

hay a=7

a: 3x^3+2x^2-7x+a chia hêt cho 3x-1

=>3x^3-x^2+3x^2-x-6x+2+a-2 chia hết cho 3x-1

=>a-2=0

=>a=2

c: =>2x^2-6x+(a+6)x-3a-18+3a+19 chia x-3 dư 4

=>3a+19=4

=>3a=-15

=>a=-5

d: 2x^3-x^2+ax+b chiahêt cho x^2-1

=>2x^3-2x-x^2+1+(a+2)x+b-1 chia hết cho x^2-1

=>a+2=0 và b-1=0

=>a=-2 và b=1

\(2x^2+mx+1=f\left(x\right).\left(x-3\right)+4\)

Thay \(x=3\)

\(\Rightarrow2.3^2+3m+1=4\)

\(\Rightarrow3m=-15\)

\(\Rightarrow m=-5\)

Bài 1:

\(2x^4+ax^2+bx+c⋮x-2\\ \Leftrightarrow2x^4+ax^2+bx+c=\left(x-2\right)\cdot a\left(x\right)\)

Thay \(x=2\Leftrightarrow32+4a+2b+c=0\Leftrightarrow4a+2b+c=-32\left(1\right)\)

\(2x^4+ax^2+bx+c:\left(x^2-1\right)R2x\\ \Leftrightarrow2x^4+ax^2+bx+c=\left(x-1\right)\left(x+1\right)\cdot b\left(x\right)+2x\)

Thay \(x=1\Leftrightarrow2+a+b+c=2\Leftrightarrow a+b+c=0\left(2\right)\)

Thay \(x=-1\Leftrightarrow2+a-b+c=-2\Leftrightarrow a-b+c=-4\left(3\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}4a+2b+c=-32\\a+b+c=0\\a-b+c=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{34}{3}\\b=2\\c=\dfrac{28}{3}\end{matrix}\right.\)

Bài 2:

Do \(f\left(x\right):x^2+x-12\) được thương bậc 2 nên dư bậc 1

Gọi đa thức dư là \(ax+b\)

Vì \(f\left(x\right):x^2+x-12\) được thương là \(x^2+3\) và còn dư nên

\(f\left(x\right)=\left(x^2+x-12\right)\left(x^2+3\right)+ax+b\\ \Leftrightarrow f\left(x\right)=\left(x+4\right)\left(x-3\right)\left(x^2+3\right)+ax+b\)

Thay \(x=3\Leftrightarrow f\left(3\right)=3a+b\)

Mà \(f\left(x\right):\left(x-3\right)R2\Leftrightarrow f\left(3\right)=2\Leftrightarrow3a+b=2\left(1\right)\)

Thay \(x=-4\Leftrightarrow f\left(-4\right)=-4a+b\)

Mà \(f\left(x\right):\left(x+4\right)R9\Leftrightarrow f\left(-4\right)=9\Leftrightarrow-4a+b=-9\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}3a+b=2\\-4a+b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=5\end{matrix}\right.\)

Do đó \(f\left(x\right)=\left(x^2+x-12\right)\left(x^2+3\right)-x+5\)

\(\Leftrightarrow f\left(x\right)=x^4+3x^2+x^3+3x-12x^2-36-x+5\\ \Leftrightarrow f\left(x\right)=x^4+x^3-9x^2+2x-31\)

Ta có x3+ax+bx3+ax+b chia x+1x+1 dư 7.
Suy ra x3+ax+b=Q(x).(x+1)+7x3+ax+b=Q(x).(x+1)+7
Với x=−1x=−1 thì f(−1)=b−a−1=7⇒b−a=8⇒b=a+8f(−1)=b−a−1=7⇒b−a=8⇒b=a+8.
Lại có x3+ax+b=H(x).(x−3)−5x3+ax+b=H(x).(x−3)−5.
Với x=3x=3 thì f(3)=27+3a+b=−5⇒3a+b=−22f(3)=27+3a+b=−5⇒3a+b=−22.
Thay vô ta Tim được a,b

Cậu chú ý mũ nha , tớ không viết kịp

Ta phân tích thành

\(x^3+ax+b=\left(x+1\right)\left(x^2-x+a+1\right)+b-a-1\)

Và \(x^3+ax+b=\left(x-2\right)\left(x^2+2x+a+4\right)+b+2a+8\)

Kết hợp với đề bài ta có hệ

\(\hept{\begin{cases}b-a-1=7\\b+2a+8=4\end{cases}\Leftrightarrow\hept{\begin{cases}a=-4\\b=4\end{cases}}}\)

\(a,\Leftrightarrow4x^3-2x^2+a=\left(2x-3\right).a\left(x\right)\)

Thay \(x=\dfrac{3}{2}\Leftrightarrow4.\dfrac{27}{8}-2.\dfrac{9}{4}+a=0\)

\(\Leftrightarrow\dfrac{27}{2}-\dfrac{9}{2}+a=0\\ \Leftrightarrow a=-9\)

\(b,\Leftrightarrow3x^3+2x^2+x+a=\left(x+1\right).b\left(x\right)+2\)

Thay \(x=-1\Leftrightarrow-3+2-1+a=2\Leftrightarrow a=4\)