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 [imath]x[/imath] and thus get my cubic solutions. Everything went smoothly until I reached a point where I have an equation of the form [imath]ax^2 + bx + c + \frac{d}{x} = 0[/imath]. This form is the quadratic reduction of the original cubic of the form [imath]ax^3 + bx^2 + cx + d = 0[/imath]. I'm wondering how I should factor this, so that I can then solve for [imath]x[/imath]. Here's what I have so far:

Solve for x

[math]x^3 - 6x^2 + 2x + 3 = 0[/math]
Subtract 3 from both sides to get

[math]x^3 - 6x^2 + 2x = -3[/math]
Multiply both sides by [imath]\frac{1}{x}[/imath] to get

[math]x^2 - 6x + 2 = \frac{-3}{x}[/math]
Add [imath]\frac{3}{x}[/imath] to both sides to get

[math]x^2 - 6x + 2 + \frac{3}{x} = 0[/math]
And here I am, stuck with this [imath]\frac{d}{x}[/imath] 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 [imath]\frac{d}{x}[/imath] term. So, how can I go about factoring this quadratic with a [imath]\frac{d}{x}[/imath] term in it? Do I have to move the [imath]\frac{d}{x}[/imath] term back to the other side and then factor [imath]x^2 - 6x + 2[/imath] as if it equals 0 and then solve for [imath]x[/imath]? 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 [imath]x[/imath] and thus get my cubic solutions. Everything went smoothly until I reached a point where I have an equation of the form [imath]ax^2 + bx + c + \frac{d}{x} = 0[/imath]. This form is the quadratic reduction of the original cubic of the form [imath]ax^3 + bx^2 + cx + d = 0[/imath]. I'm wondering how I should factor this, so that I can then solve for [imath]x[/imath]. Here's what I have so far:

Solve for x

[math]x^3 - 6x^2 + 2x + 3 = 0[/math]
Subtract 3 from both sides to get

[math]x^3 - 6x^2 + 2x = -3[/math]
Multiply both sides by [imath]\frac{1}{x}[/imath] to get

[math]x^2 - 6x + 2 = \frac{-3}{x}[/math]
Add [imath]\frac{3}{x}[/imath] to both sides to get

[math]x^2 - 6x + 2 + \frac{3}{x} = 0[/math]
And here I am, stuck with this [imath]\frac{d}{x}[/imath] 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 [imath]\frac{d}{x}[/imath] term. So, how can I go about factoring this quadratic with a [imath]\frac{d}{x}[/imath] term in it? Do I have to move the [imath]\frac{d}{x}[/imath] term back to the other side and then factor [imath]x^2 - 6x + 2[/imath] as if it equals 0 and then solve for [imath]x[/imath]? 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 [imath]x[/imath] 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 [imath]ax^2[/imath]? That holds true for my expression of [imath]x^2 - 6x + 2 + \frac{3}{x} = 0[/imath] that I got from dividing every term of the cubic by [imath]x[/imath].

 

[Deleted member 4993]

Wait, why can’t I just divide every term of the cubic by [imath]x[/imath] 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 [imath]ax^2[/imath]? That holds true for my expression of [imath]x^2 - 6x + 2 + \frac{3}{x} = 0[/imath] that I got from dividing every term of the cubic by [imath]x[/imath].

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)

[imath]x^2 - 6x + 2 + \frac{3}{x} = x^2 - 6x + 2 + 3 x^{-1} = 0[/imath] 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 [imath]x[/imath] and thus get my cubic solutions. Everything went smoothly until I reached a point where I have an equation of the form [imath]ax^2 + bx + c + \frac{d}{x} = 0[/imath]. This form is the quadratic reduction of the original cubic of the form [imath]ax^3 + bx^2 + cx + d = 0[/imath]. I'm wondering how I should factor this, so that I can then solve for [imath]x[/imath]. Here's what I have so far:

Solve for x

[math]x^3 - 6x^2 + 2x + 3 = 0[/math]
Subtract 3 from both sides to get

[math]x^3 - 6x^2 + 2x = -3[/math]
Multiply both sides by [imath]\frac{1}{x}[/imath] to get

[math]x^2 - 6x + 2 = \frac{-3}{x}[/math]
Add [imath]\frac{3}{x}[/imath] to both sides to get

[math]x^2 - 6x + 2 + \frac{3}{x} = 0[/math]
And here I am, stuck with this [imath]\frac{d}{x}[/imath] 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 [imath]\frac{d}{x}[/imath] term. So, how can I go about factoring this quadratic with a [imath]\frac{d}{x}[/imath] term in it? Do I have to move the [imath]\frac{d}{x}[/imath] term back to the other side and then factor [imath]x^2 - 6x + 2[/imath] as if it equals 0 and then solve for [imath]x[/imath]? 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 [imath]x[/imath] 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 [imath]ax^2[/imath]? That holds true for my expression of [imath]x^2 - 6x + 2 + \frac{3}{x} = 0[/imath] that I got from dividing every term of the cubic by [imath]x[/imath].

I love you man!!!

Cubic equation - Wikipedia

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