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Sử dụng định lý Bezout:
a/ \(g\left(x\right)=0\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(f\left(x\right)⋮g\left(x\right)\Rightarrow\left\{{}\begin{matrix}f\left(1\right)=0\\f\left(2\right)=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=1\\2a+b=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3\\b=-2\end{matrix}\right.\)
b/ \(g\left(x\right)=0\Rightarrow x=-1\)
\(\Rightarrow f\left(-1\right)=0\Rightarrow-a+b=2\Rightarrow b=a+2\)
Tất cả các đa thức có dạng \(f\left(x\right)=2x^3+ax+a+2\) đều chia hết \(g\left(x\right)=x+1\) với mọi a
c/ \(g\left(x\right)=0\Rightarrow x=-2\Rightarrow f\left(-2\right)=0\Rightarrow4a+b=-30\)
\(2x^4+ax^2+x+b=\left(x^2-1\right).Q\left(x\right)+x\)
Thay \(x=1\Rightarrow a+b=-2\)
\(\Rightarrow\left\{{}\begin{matrix}4a+b=-30\\a+b=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\frac{28}{3}\\b=\frac{22}{3}\end{matrix}\right.\)
d/ Tương tự: \(\left\{{}\begin{matrix}f\left(2\right)=8a+4b-40=0\\f\left(-5\right)=-125a+25b-75=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\\b=\end{matrix}\right.\)
a) Ta có: \(g\left(x\right)=x^2-3x+2\)
\(=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
Vì \(f\left(x\right)⋮g\left(x\right)\)
\(\Rightarrow f\left(x\right)=\left(x-1\right)\left(x-2\right)q\left(x\right)\)
\(\Rightarrow\hept{\begin{cases}f\left(1\right)=\left(1-1\right)\left(1-2\right)q\left(1\right)=0\left(1\right)\\f\left(2\right)=\left(1-2\right)\left(2-2\right)q\left(2\right)=0\left(2\right)\end{cases}}\)
Từ \(\left(1\right)\Leftrightarrow1^4-3.1^3+1^2+a+b=0\)
\(\Leftrightarrow-1+a+b=0\)
\(\Leftrightarrow a+b=1\left(3\right)\)
Từ \(\left(2\right)\Leftrightarrow2^4-3.2^3+2^2+2a+b=0\)
\(\Leftrightarrow-4+2a+b=0\)
\(\Leftrightarrow2a+b=4\left(4\right)\)
Từ \(\left(3\right);\left(4\right)\Rightarrow\hept{\begin{cases}a+b=1\\2a+b=4\end{cases}\Leftrightarrow\hept{\begin{cases}a=3\\b=-2\end{cases}}}\)
Vậy a=3 và b=-2 để \(f\left(x\right)⋮g\left(x\right)\)
Các phần sau tương tự
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a) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(x^3-3x+2=x^3+2x^2-2x^2-4x+x+2\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+1\right)=\left(x+2\right)\left(x-1\right)^2\)
c) \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+9\right)=\left(x+1\right)\left(x-3\right)^2\)
d) \(x^3+8x^2+17x+10=x^3+2x^2+6x^2+12x+5x+10\)
\(=x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+6x+5\right)=\left(x+2\right)\left(x+5\right)\left(x+1\right)\)
e) \(x^3+3x^2+6x+4=x^3+x^2+2x^2+2x+4x+4\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
f) \(x^3+3x^2+3x+2=x^3+2x^2+x^2+2x+x+2\)
\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+x+1\right)\)
1, \(\left(2x^4-5x^2y^2+3xy^3\right)\left(5x^3+x^2y-y^3\right)\)
\(=10x^7-25x^5y^2+15x^4y^3+2x^6y-5x^4y^3+5x^2y^5+3xy^6\)
2, a, \(4-2x+5x^2-4x^2\&5x-3+x^2\)
Sắp xếp: \(4-2x+5x^2-4x^2=5x^2-4x^2-2x+4=x^2-2x+4\)
\(5x-3+x^2=x^2+5x-3\)
- \(\left(x^2-2x+4\right)\left(x^2+5x-3\right)=x^4+3x^3-9x^2-14x-12\)
b, Làm tương tự câu a
1 ) \(\left(2x^4-5x^2y^2+3xy^3\right)\left(5x^3+x^2y-y^3\right)\)
\(=2x^4\left(5x^3+x^2y-y^3\right)-5x^2y^2\left(5x^3+x^2y-y^3\right)+3xy^3\left(5x^3+x^2y-y^3\right)\)\(=10x^7+2x^6y-2x^4y-25x^5y^2-5x^4y^3+5x^2y^5+15x^4y^3+3x^3y^4-3xy^6\)2 ) a ) \(4-2x+5x^2-4x^2=x^2-2x+4\)
\(5x-3+x^2=x^2+5x-3\)
\(\left(x^2-2x+4\right)\left(x^2+5x-3\right)\)
\(=x^4-2x^3+4x^2+5x^3-10x^2+20x-3x^2+6x-12\)
\(=x^4+3x^3-9x^2+26x-12\)
b ) \(10-x^4+3x-4x^2=-x^4-4x^2+3x+10\)
\(2x+x^3-1=x^3+2x-1\)
\(\left(-x^4-4x^2+3x+10\right)\left(x^3+2x-1\right)\)
\(=-x^4\left(x^3+2x-1\right)-4x^2\left(x^3+2x-1\right)+3x\left(x^3+2x-1\right)+10\left(x^3+2x-1\right)\)\(=-x^7-2x^5+x^4-4x^5-8x^3+4x^2+3x^4+6x^2-3x+10x^3+20x-10\)\(=-x^7-\left(2x^5+4x^5\right)+\left(3x^4+x^4\right)+\left(10x^3-8x^3\right)+\left(4x^2+6x^2\right)+\left(20x-3x\right)-10\)\(=-x^7-6x^5+4x^4+2x^3+10x^2+17x-10\)
Bài 1 :
Gọi f( x ) = 2n2 + n - 7
g( x ) = n - 2
Cho g( x ) = 0
\(\Leftrightarrow\)n - 2 = 0
\(\Rightarrow\)n = 2
\(\Leftrightarrow\)f( 2 ) = 2 . 22 + 2 - 7
\(\Rightarrow\)f( 2 ) = 3
Để f( x ) \(⋮\)g( x )
\(\Rightarrow\)n - 2 \(\in\)Ư( 3 ) = { \(\pm\)1 ; \(\pm\)3 }
Ta lập bảng :
| n - 2 | 1 | - 1 | 3 | - 3 |
| n | 3 | 1 | 5 | - 1 |
Vậy : n \(\in\){ - 1 ; 1 ; 3 ; 5 }