How can i factor (with long divison) a polynomial that doesn't have a constant?

[abel muroi]

I was given the polynomial and was told to put it in factored form..

2x³ + 5x² + 6x

I'm trying to factor this polynomial with the long division method, but i am having a bit of trouble doing that because the polynomial doesn't have a constant at the end (...6x + 12)

So my question is, how can I factor this properly? I'm so used to factoring the polynomials that have a constant at the end..

 

[Quaid]

Factor out an x, first.

The resulting quadratic-factor's discriminant (i.e., B^2-4AC) is negative; this tells us that the polynomial does not factor further, within the Real number system.

Try the above with this one, instead.

2x^3 + 5x^2 + 2x

 

[Deleted member 4993]

I was given the polynomial and was told to put it in factored form..

2x³ + 5x² + 6x

I'm trying to factor this polynomial with the long division method, but i am having a bit of trouble doing that because the polynomial doesn't have a constant at the end (...6x + 12)

So my question is, how can I factor this properly? I'm so used to factoring the polynomials that have a constant at the end..

2x³ + 5x² + 6x = x ( 2x² + 5x + 6) .... now you have the constant you were looking for. Analyze the function inside the parenthesis () for further factorization.