Phân tích đa thức thành nhân tử:x^2+3x+2
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\(x^3-3x^2+1-3x=\left(x^3+1\right)-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1-3x\right)=\left(x+1\right)\left(x^2-4x+1\right)\)
\(x^8+3x^3+1\)
\(=x^8-x^4+4x^4+4\)
\(=\left(x^4-1\right)\cdot\left(x^4+1\right)+4\cdot\left(x^4+1\right)\)
\(=\left(x^4+1\right)\cdot\left(x^4-1+4\right)\)
\(=\left(x^4+1\right)\cdot\left(x^4+3\right)\)
\(=x^2+x+2x+2=x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(x+2\right)\)
\(x^4-x^2+2x+2\)
\(=x^4-2x^3+2x^2+2x^3-4x^2+4x+x^2-2x+2\)
\(=\left(x^4-2x^3+2x^2\right)+\left(2x^3-4x^2+4x\right)+\left(x^2-2x+2\right)\)
\(=x^2\left(x^2-2x+2\right)+2x\left(x^2-2x+2\right)+\left(x^2-2x+2\right)\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+1\right)\)
\(=\left(x^2-2x+2\right)\left(x+1\right)^2\)
\(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
\(\dfrac{1}{4}x^2+2xy+4y^2=\left(\dfrac{1}{2}x+2y\right)^2\)
\(=x^2+2x\cdot\frac{1}{2}+\frac{1}{4}-\left(\frac{\sqrt{23}}{2}i\right)^2\)
\(=\left(x+\frac{1}{2}\right)^2\)\(-\left(\frac{\sqrt{23}}{2}i\right)^2\)
\(\left(x+\frac{1}{2}-\frac{\sqrt{23}}{2}i\right)\left(x+\frac{1}{2}+\frac{\sqrt[]{23}}{2}i\right)\)
\(x^2+3x+2=x^2+2x+x+2\)
\(=x\left(x+1\right)+2\left(x+1\right)=\left(x+2\right)\left(x+1\right)\)