Real Solutions for a Complex System

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SUMMARY

The discussion focuses on solving the polynomial equation $(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24=0$. Participants emphasize the importance of factoring each quadratic component to simplify the equation. The factors include $(x+1)(x+2)$, $(x-1)(x+2)$, and $(x-3)(x-4)$, leading to the identification of real solutions. The final solutions are derived from setting each factor equal to zero and solving for x.

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  • Understanding of polynomial equations
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  • Familiarity with the zero-product property
  • Basic algebraic manipulation skills
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  • Learn about the Rational Root Theorem
  • Explore synthetic division for polynomial equations
  • Investigate graphical methods for finding polynomial roots
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Students, educators, and mathematics enthusiasts interested in solving polynomial equations and enhancing their algebraic skills.

anemone
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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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