MHB Real Solutions for a Complex System

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Solve for all real solutions for the system below:

$(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24=0$
 
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anemone said:
Solve for all real solutions for the system below:

$(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24=0$

$$(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24$$

$$=x^6-6x^5+40x^3-13x^2-70x$$

$$=x(x^5-6x^4+40x^2-13x-70)$$

$$=x(x-2)(x^4-4x^3-8x^2+24x+35)$$

$$=x(x-2)(x^2-2x-5)(x^2-2x-7)=0$$

$$\implies x\in\left\{0,2,1\pm\sqrt6,1\pm2\sqrt2\right\}$$
 
greg1313 said:
$$(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24$$

$$=x^6-6x^5+40x^3-13x^2-70x$$

$$=x(x^5-6x^4+40x^2-13x-70)$$

$$=x(x-2)(x^4-4x^3-8x^2+24x+35)$$

$$=x(x-2)(x^2-2x-5)(x^2-2x-7)=0$$

$$\implies x\in\left\{0,2,1\pm\sqrt6,1\pm2\sqrt2\right\}$$

Sorry greg1313 for the late reply!

Very well done greg1313! And thanks for participating!
 
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