How do I factor a quadratic with a rational expression of x in it?

[caters]

So, the thought came to me of solving a cubic equation. I've done it a few times before, but it has been years. I know how to solve quadratics, so I was thinking that if I could reduce the cubic to a quadratic, then I could solve the quadratic for $x$ and thus get my cubic solutions. Everything went smoothly until I reached a point where I have an equation of the form $ax^2 + bx + c + \frac{d}{x} = 0$. This form is the quadratic reduction of the original cubic of the form $ax^3 + bx^2 + cx + d = 0$. I'm wondering how I should factor this, so that I can then solve for $x$. Here's what I have so far:

Solve for x

$$x^3 - 6x^2 + 2x + 3 = 0$$
Subtract 3 from both sides to get

$$x^3 - 6x^2 + 2x = -3$$
Multiply both sides by $\frac{1}{x}$ to get

$$x^2 - 6x + 2 = \frac{-3}{x}$$
Add $\frac{3}{x}$ to both sides to get

$$x^2 - 6x + 2 + \frac{3}{x} = 0$$
And here I am, stuck with this $\frac{d}{x}$ term in my quadratic and not knowing how to factor a quadratic with such a term in it. When I try to find an answer to my question, all I get is how to factor a quadratic that has fractional coefficients, which is not at all the same as factoring a quadratic with a $\frac{d}{x}$ term. So, how can I go about factoring this quadratic with a $\frac{d}{x}$ term in it? Do I have to move the $\frac{d}{x}$ term back to the other side and then factor $x^2 - 6x + 2$ as if it equals 0 and then solve for $x$? Or is there another way to go about it that keeps the right side equal to 0?

 

[Deleted member 4993]

So, the thought came to me of solving a cubic equation. I've done it a few times before, but it has been years. I know how to solve quadratics, so I was thinking that if I could reduce the cubic to a quadratic, then I could solve the quadratic for $x$ and thus get my cubic solutions. Everything went smoothly until I reached a point where I have an equation of the form $ax^2 + bx + c + \frac{d}{x} = 0$. This form is the quadratic reduction of the original cubic of the form $ax^3 + bx^2 + cx + d = 0$. I'm wondering how I should factor this, so that I can then solve for $x$. Here's what I have so far:

Solve for x

$$x^3 - 6x^2 + 2x + 3 = 0$$
Subtract 3 from both sides to get

$$x^3 - 6x^2 + 2x = -3$$
Multiply both sides by $\frac{1}{x}$ to get

$$x^2 - 6x + 2 = \frac{-3}{x}$$
Add $\frac{3}{x}$ to both sides to get

$$x^2 - 6x + 2 + \frac{3}{x} = 0$$
And here I am, stuck with this $\frac{d}{x}$ term in my quadratic and not knowing how to factor a quadratic with such a term in it. When I try to find an answer to my question, all I get is how to factor a quadratic that has fractional coefficients, which is not at all the same as factoring a quadratic with a $\frac{d}{x}$ term. So, how can I go about factoring this quadratic with a $\frac{d}{x}$ term in it? Do I have to move the $\frac{d}{x}$ term back to the other side and then factor $x^2 - 6x + 2$ as if it equals 0 and then solve for $x$? Or is there another way to go about it that keeps the right side equal to 0?

You CANNOT reduce a cubic polynomial to quadratic, by that method.
Google
Cardano's cubic solution.

 

[Harry_the_cat]

Your expression which involves 3/x is NOT a quadratic.

Have you heard of the "Factor Theorem"? This states that:
If f(x) is a polynomial and f(c)=0 then (x−c) is a factor of f(x).

Note that if we call your cubic function f(x), then f(1) =0 [ "How did I get 1" I hear you say. This is an educated "guess" by considering the coefficients].

f(1) = 0 means that (x - 1) is a factor.

Do you know how to do polynomial long division to get the other factor (the other factor will be a quadratic)? Or perhaps another method to get the other factor?

 

[caters]

You CANNOT reduce a cubic polynomial to quadratic, by that method.
Google
Cardano's cubic solution.

Wait, why can’t I just divide every term of the cubic by $x$ to reduce it to a quadratic? It seems like it should be possible to do that.

Your expression which involves 3/x is NOT a quadratic.

Why not? Isn’t a quadratic an expression with a maximum of degree 2, thus the highest degree term being $ax^2$? That holds true for my expression of $x^2 - 6x + 2 + \frac{3}{x} = 0$ that I got from dividing every term of the cubic by $x$.

 

[Deleted member 4993]

Wait, why can’t I just divide every term of the cubic by $x$ to reduce it to a quadratic? It seems like it should be possible to do that.

Why not? Isn’t a quadratic an expression with a maximum of degree 2, thus the highest degree term being $ax^2$? That holds true for my expression of $x^2 - 6x + 2 + \frac{3}{x} = 0$ that I got from dividing every term of the cubic by $x$.

Definition of polynomial (Meriam-Webster):

Definition of polynomial​

(Entry 1 of 2)
: a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a nonnegative integral power (such as a + bx + cx²)

$x^2 - 6x + 2 + \frac{3}{x} = x^2 - 6x + 2 + 3 x^{-1} = 0$ is NOT a quadratic polynomial

 

[jonah2.0]

Beer induced declaration follows.

So, the thought came to me of solving a cubic equation. I've done it a few times before, but it has been years. I know how to solve quadratics, so I was thinking that if I could reduce the cubic to a quadratic, then I could solve the quadratic for $x$ and thus get my cubic solutions. Everything went smoothly until I reached a point where I have an equation of the form $ax^2 + bx + c + \frac{d}{x} = 0$. This form is the quadratic reduction of the original cubic of the form $ax^3 + bx^2 + cx + d = 0$. I'm wondering how I should factor this, so that I can then solve for $x$. Here's what I have so far:

Solve for x

$$x^3 - 6x^2 + 2x + 3 = 0$$
Subtract 3 from both sides to get

$$x^3 - 6x^2 + 2x = -3$$
Multiply both sides by $\frac{1}{x}$ to get

$$x^2 - 6x + 2 = \frac{-3}{x}$$
Add $\frac{3}{x}$ to both sides to get

$$x^2 - 6x + 2 + \frac{3}{x} = 0$$
And here I am, stuck with this $\frac{d}{x}$ term in my quadratic and not knowing how to factor a quadratic with such a term in it. When I try to find an answer to my question, all I get is how to factor a quadratic that has fractional coefficients, which is not at all the same as factoring a quadratic with a $\frac{d}{x}$ term. So, how can I go about factoring this quadratic with a $\frac{d}{x}$ term in it? Do I have to move the $\frac{d}{x}$ term back to the other side and then factor $x^2 - 6x + 2$ as if it equals 0 and then solve for $x$? Or is there another way to go about it that keeps the right side equal to 0?

Wait, why can’t I just divide every term of the cubic by $x$ to reduce it to a quadratic? It seems like it should be possible to do that.

Why not? Isn’t a quadratic an expression with a maximum of degree 2, thus the highest degree term being $ax^2$? That holds true for my expression of $x^2 - 6x + 2 + \frac{3}{x} = 0$ that I got from dividing every term of the cubic by $x$.

I love you man!!!

Cubic equation - Wikipedia

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