phân tích đa thức thành nhân tử : \(2x^3-x^2+5x+3\) -
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$ 2x^3 - x^2 + 5x + 3 \\ = 2x^3 + x^2 - 2x^2 - x + 6x + 3 \\ = x^2(2x + 1) - x(2x + 1) + 3(2x + 1) \\ = (2x + 1)(x^2 - x + 3) $
\(2x^3-x^2+5x+3\)
= \(2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
Vì \(x^2-x+3=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+3>0\)
Nên
\(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
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\(x^3+x+2=\left(x^3+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
\(b,x^4+5x^3+10x-4=\left(x^4-4\right)+\left(5x^3-10x\right)\)\(=\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2-2+5x\right)\)
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x4+x3+2x2+x+1=x4+x3+x2+x2+x+1=(x4+x3+x2)+(x2+x+1)
=x2(x2+x+1)+(x2+x+1)
=(x2+x+1)(x2+1)
=(x^4+2x^2+1)+(x^3+x)
=(x^2+1)^2+x(x^2+1)
(x^+1)*(x^2+1+x0
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2x3-x2+5x+3=2x3+x2-2x2-x+6x+3=(2x3+x2)-(2x2+x)+(6x+3)=x2(2x+1)-x(2x+1)+3(2x+1)=(2x+1)(x2-x+3)
#)Giải :
\(2x^3-x^2+5x+3\)
\(=2x^3-2x^2+x^2+6x-x+3\)
\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)