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\frac{x^4+5x^3-1\cdot 20\cdot x-16}{x^3+2x^2-13\cdot x+10}\leq 0

Solve the inequality (20x*-1-16+5x^3+x^4)/(-13x+10+2x^2+x^3)#0

Answer

$\frac{\left(1+x\right)\left(x-2\right)\left(x^{2}+6x+8\right)}{x\left(x^{2}+3x-10\right)-\left(x^{2}+3x-10\right)}\leq 0$

Step-by-step explanation

Problem

$\frac{x^4+5x^3-1\cdot 20\cdot x-16}{x^3+2x^2-13\cdot x+10}\leq 0$
1

Multiply $-1$ times $20$

$\frac{-16-20x+5x^3+x^4}{10-13x+2x^2+x^3}\leq 0$

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Answer

$\frac{\left(1+x\right)\left(x-2\right)\left(x^{2}+6x+8\right)}{x\left(x^{2}+3x-10\right)-\left(x^{2}+3x-10\right)}\leq 0$

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