Phân tích đa thức : \(x^2+9x+3\)
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x^4 - 9x^3 + x^2 - 9x
= x ( x^3 - 9x^2 + x - 9 )
= x [ x( x^2 +1 ) - 9 ( x^2 + 1 )]
= x ( x^2 + 1 ( x - 9 )
\(x^3-9x^2+x=x\left(x^2-9x+1\right)\)
\(x^3+13x^2+x=x\left(x^2+13x+1\right)\)
a) \(27+x^3=\left(x+3\right)\left(x^2-3x+9\right)\)
b) \(-x^3+12x^2-48x+64=\left(4-x\right)^3\)
c) \(27+27x+9x^2=9\left(x^2+3x+3\right)\)
\(=x^3+2x^2-8x=x\left(x^2+2x-8\right)\\ =x\left(x^2-2x+4x-8\right)\\ =x\left(x-2\right)\left(x+4\right)\)
\(x^3-9x^2+14x\)
= \(x^3-7x^2-2x^2+14x\)
= \(x^2.\left(x-7\right)-2x.\left(x-7\right)\)
= \(\left(x-7\right).\left(x^2-2x\right)\)
= \(\left(x-7\right).\left(x-2\right).x\)
\(x^3-9x^2+14x\)
\(=x\left(x^2-9x+14\right)\)
\(=x\left(x^2-7x-2x+14\right)\)
\(=x\left[x\left(x-7\right)-2\left(x-7\right)\right]\)
\(=x\left(x-2\right)\left(x-7\right)\)
\(27+27x+9x^2+x^3\)
\(=x^3+9x^2+27x+27\)
\(=x^3+3x^2+6x^2+18x+9x+27\)
\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+6x+9\right)\)
\(=\left(x+3\right)\left(x+3\right)^2=\left(x+3\right)^3\)
Nếu nhìn kĩ thì bạn sẽ thấy đây là hằng đẳng thức nhé !
\(x^3+9x^2+27x+27=x^3+3.3.x^2+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
\(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(x-2\right).\)
\(x^2\left(x-2\right)-4x\left(x-2\right)+\left(x-2\right)\)
vậy................
\(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(x-2\right)\)
\(x^2\left(x-2\right)-4x\left(x-2\right)+\left(x-2\right)\)
Vậy ........
x2 + 9x + 3 = x2 + 6x + 3x + 3
= (x+3)2 + 3x
Phân tích đa thức thành nhân tử :
\(x^2+9x+3=x^2+2.x.\frac{9}{2}+\left(\frac{9}{2}\right)^2-\frac{81}{4}+3\)
\(=\left(x+\frac{9}{2}\right)^2-\frac{69}{4}=\left(x+\frac{9}{2}\right)^2-\left(\frac{\sqrt{69}}{2}\right)^2\)
\(=\left(x+\frac{9}{2}-\frac{\sqrt{69}}{2}\right)\left(x+\frac{9}{2}+\frac{\sqrt{69}}{2}\right)\)
\(=\left(x+\frac{9-\sqrt{69}}{2}\right)\left(x+\frac{9+\sqrt{69}}{2}\right)\)