Phân tích đa thức thành nhân tử:
a. \(x^3+8x^2+14x+10\)
b. \(x^4+x^3+2x^2+x+1\)
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\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)
a: \(x^4-2x^3+x^2-2x\)
\(=\left(x^4-2x^3\right)+\left(x^2-2x\right)\)
\(=x^3\left(x-2\right)+x\left(x-2\right)\)
\(=x\left(x-2\right)\left(x^2+1\right)\)
b: \(x^4+x^3-8x-8\)
\(=\left(x^4+x^3\right)-\left(8x+8\right)\)
\(=x^3\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-8\right)\)
\(=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right).\left[x^2+4x-3x-12\right]\)
\(=\left(x-2\right).\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
\(x^4+x^3+2x-4\)
\(=x^4-x^3+2x^3-2x^2+2x^2-2x+4x-4\)
\(=x^3\left(x-1\right)+2x^2\left(x-1\right)+2x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+2x+4\right)\)
\(=\left(x-1\right).\left[x^2\left(x+2\right)+2\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+2\right)\)
\(8x^4-2x^3-3x^2-2x-1\)
\(=8x^4-8x^3+6x^3-6x^2+3x^2-3x+x-1\)
\(=8x^3\left(x-1\right)+6x^2\left(x-1\right)+3x\left(x-1\right)+x-1\)
\(=\left(x-1\right)\left(8x^3+6x^2+3x+1\right)\)
\(=\left(x-1\right)\left[\left(8x^3+1\right)+\left(6x^2+3x\right)\right]\)
\(=\left(x-1\right)\left[\left(2x+1\right)\left(4x^2-2x+1\right)+3x\left(2x+1\right)\right]\)
\(=\left(x-1\right)\left(2x+1\right)\left(4x^2+x+1\right)\)
\(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
Chúc bạn học tốt.
Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
\(x^4+x^3+2x^2+x+1\)
\(=x^4+x^3+x^2+x^2+x+1\)
\(=\left(x^4+x^2\right)+\left(x^3+x\right)+\left(x^2+1\right)\)
\(=x^2\left(x^2+1\right)+x\left(x^2+1\right)+1\left(x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+1\right)\)
a) Giả sử : x3 + 8x2 + 14x + 10 = ( ax + d )( x2 + bx + c)
x3 + 8x2 + 14x + 10 = ax3 + abx2 + acx + dx2 + bdx + dc
x3 + 8x2 + 14x + 10 = ax3 + x2( ab + d) + x( ac + bd) + dc
Đến đây , bạn đồng nhất đa tức là tìm ra nhé , mk đi học rùi