$$-\frac{4}{x}+\frac{3}{x^2}+1 = 0$$

Answer

$$ \begin{matrix}x_1 = 1 & x_2 = -1+\sqrt{ 2 }i & x_3 = -1-\sqrt{ 2 }i \end{matrix} $$

Explanation

$$ \begin{aligned} -\frac{4}{x}+\frac{3}{x^2}+1 &= 0&& \text{multiply ALL terms by } \color{blue}{ xx^2 }. \\[1 em]-xx^2\cdot\frac{4}{x}+xx^2\cdot\frac{3}{x^2}+xx^2\cdot1 &= xx^2\cdot0&& \text{cancel out the denominators} \\[1 em]-4+\frac{3}{x^1}+x^3 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]-x^1\cdot4+x^1\cdot\frac{3}{x^1}+x^1\cdot1x^3 &= x^1\cdot0&& \text{cancel out the denominators} \\[1 em]-4x+3+x^4 &= 0&& \text{simplify left side} \\[1 em]x^4-4x+3 &= 0&& \\[1 em] \end{aligned} $$

$ \color{blue}{ x^{4}-4x+3 } $ is a polynomial of degree 4. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (3) are 1 3 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(1) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{x - 1} $

$$ \frac{ x^{4}-4x+3 }{ \color{blue}{ x - 1 } } = x^{3}+x^{2}+x-3 $$

Polynomial $ x^{3}+x^{2}+x-3 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{3}+x^{2}+x-3 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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