Zoe Oneal

2020-12-13

We need to solve this equation:
${x}^{4}-{x}^{3}+4{x}^{2}-16x-192=0$

### Answer & Explanation

Bentley Leach

Use the rational root theorem, ${a}_{0}=192$ and
${a}_{n}=1.$
The divider of ${a}_{0}is1,2,3,4,6,8,12,16,24,32,48,96,192.$
Rational Zero $=±\frac{1,2,3,4,6,8,12,16,24,32,48,96,192}{1}$
The root of the equation is - 3 by trial and error method then, the factor is $x+3$. Now, divide the provided equation with calculated factor.
${x}^{4}-{x}^{3}+4{x}^{2}-16x-192=\frac{{x}^{4}-{x}^{3}+4{x}^{2}-16x-192}{x+3}$
$=\left(x+3\right)\left({x}^{3}-4x2+16x-64\right)$
Now, further factorize the above equation.
$\left(x+3\right)\left({x}^{3}-4{x}^{2}+16x-64\right)=0$
$\left(x+3\right)\left(\left({x}^{3}-4{x}^{2}\right)+\left(16x-64\right)\right)=0$
$\left(x+3\right)\left({x}^{2}\left(x-4\right)+16\left(x-4\right)\right)=0$
$\left(x+3\right)\left(x-4\right)\left({x}^{2}+16\right)=0$
$x=-3,4,±4i$
Answer: Therefore, the real solution of the provided equation is $x=-3,4.$

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