Zoe Oneal

2020-12-13

We need to solve this equation:

${x}^{4}-{x}^{3}+4{x}^{2}-16x-192=0$

Bentley Leach

Skilled2020-12-14Added 109 answers

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$=\pm \frac{1,2,3,4,6,8,12,16,24,32,48,96,192}{1}$

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.

${x}^{4}-{x}^{3}+4{x}^{2}-16x-192=\frac{{x}^{4}-{x}^{3}+4{x}^{2}-16x-192}{x+3}$

$=(x+3)({x}^{3}-4x2+16x-64)$

Now, further factorize the above equation.

$(x+3)({x}^{3}-4{x}^{2}+16x-64)=0$

$(x+3)(({x}^{3}-4{x}^{2})+(16x-64))=0$

$(x+3)({x}^{2}(x-4)+16(x-4))=0$

$(x+3)(x-4)({x}^{2}+16)=0$

$x=-3,4,\pm 4i$

Answer: Therefore, the real solution of the provided equation is$x=-3,4.$

The divider of

Rational Zero

The root of the equation is - 3 by trial and error method then, the factor is

Now, further factorize the above equation.

Answer: Therefore, the real solution of the provided equation is

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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