a, \(t\left(t+2a^2\right)+a^4=t^2+2a^2t+a^4=\left(a^2+t\right)^2\)
b, \(x^2+3x+2=x^2+2x+x+2=x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+1\right)\left(x+2\right)\)
c, \(x^4+5x^3+9x^2+7x+2\)
\(=x^4+x^3+4x^3+4x^2+5x^2+5x+2x+2\)
\(=x^3\left(x+1\right)+4x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x^3+4x^2+5x+2\right)\left(x+1\right)\)
\(=\left(x^3+x^2+3x^2+3x+2x+2\right)\left(x+1\right)\)
\(=\left[x^2\left(x+1\right)+3x\left(x+1\right)+2\left(x+1\right)\right]\left(x+1\right)\)
\(=\left(x^2+3x+2\right)\left(x+1\right)^2\)
\(=\left(x+2\right)\left(x+1\right)^3\)
a) \(t\left(t+2a^2\right)+a^4=t^2+2a^2+a^4=\left(a^2+t\right)^2\)
b)\(x^2+3x+2=x^2+x+2x+2=\left(x^2+x\right)+\left(2x+2\right)=x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(x+2\right)\)
c) \(x^4+5x^3+9x^2+7x+2\)
=\(x^4+x^3+4x^3+4x^2+5x^2+5x+2x+2\)
=\(\left(x^4+x^3\right)+\left(4x^3+4x^2\right)+\left(5x^2+5x\right)+\left(2x+2\right)\)
=\(x^3\left(x+1\right)+4x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)\)
=\(\left(x+1\right)\left(x^3+4x^2+5x+2\right)\)
\(\text{a) }t\left(t+2a^2\right)+a^4\\ \\=t^2+2a^2t+a^4\\ \\=\left(t+a^2\right)^2\)
\(\text{b) }x^2+3x+2\\ \\=x^2+2x+x+2\\ =\left(x^2+2x\right)+\left(x+2\right)\\=x\left(x+2\right)+\left(x+2\right)\\ \\=\left(x+1\right)\left(x+2\right)\\ \)
\(\text{c) }x^4+5x^3+9x^2+7x+2\\ \text{Do đa thức trên là đa thức bậc }4\\ \text{ nên khi phân tích thành nhân tử sẽ có dạng : }\\ \left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ \\ =x^4+cx^3+dx^2\: +ax^3+acx^2+adx+bx^2+bcx+bd\\ \\ =x^4+\left(c+a\right)x^3+\left(d+ac+b\right)x^2+\left(ad+bc\right)x+bd\\ \text{Đồng nhất đa thức trên với đa thức đã cho. }\\ \text{ Ta được: }\left\{{}\begin{matrix}c+a=5\\d+ac+b=9\\ad+bc=7\\bd=2\Leftrightarrow b=1;d=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+c=5\\ac=6\\2a+b=7\end{matrix}\right.\Leftrightarrow a=2;c=3\\ \text{Từ }a=2;b=1;c=3;d=2\text{ suy ra : }\\ x^4+5x^3+9x^2+7x+2\\ =\left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ \\ =\left(x^2+2x+1\right)\left(x^2+3x+2\right)\\ =\left(x+1\right)^2\left(x^2+2x+x+2\right)\\ \\ =\left(x+1\right)^2\left[\left(x^2+2x\right)+\left(x+2\right)\right]\\ \\ =\left(x+1\right)^2\left[x\left(x+2\right)+\left(x+2\right)\right]\\ \\ =\left(x+1\right)^2\left(x+1\right)\left(x+2\right)\\ =\left(x+1\right)^3\left(x+2\right)\)
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