Question

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

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

2020-12-14
Use the rational root theorem, $$a_{0} = 192$$ and
$$a_{n} = 1.$$
The divider of $$a_{0} is 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96, 192.$$
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.
$$x^{4} - x^{3} + 4x^{2} - 16x - 192 = \frac{x^{4}-x^{3}+4x^{2}-16x-192}{x+3}$$
$$= (x + 3)(x^{3} - 4x2 + 16x - 64)$$
Now, further factorize the above equation.
$$(x + 3)(x^{3} - 4x^{2} + 16x - 64) = 0$$
$$(x + 3)((x^{3} - 4x^{2}) + (16x - 64)) = 0$$
$$(x + 3)(x^{2}(x - 4) + 16(x - 4)) = 0$$
$$(x + 3)(x - 4)(x^{2} + 16) = 0$$
$$x = -3, 4, \pm 4i$$
Answer: Therefore, the real solution of the provided equation is $$x = -3, 4.$$