How Do You Factor the Numerator of a Rational Function to Find Intercepts?

[math4life]

Homework Statement

Y₁=(x³+3x²-4)/(x²)

Homework Equations

Slant Asymptote at y=x+3. (X-1) and (X+2) appear to be intercepts in the back of the book. How do I factor the numerator to get that?

The Attempt at a Solution

I know this is simple and there is a method to find the zeros of the numerator that I am overlooking- please help.

 

[symbolipoint]

A more systematic method is to apply Rational Roots Theorem. You want to find some binomials that can divide the polynomial numerator and leave no remainder. Try dividing by (X - 1) and see what results. Can you factor this result? OR, try dividing the numerator by (X + 2). How is the result? Remainder?

My guess is you want three linear binomials as a factorization for the numerator, since it has degree of 3. If those other binomials, X-1 and X+2 are factors, then your function would have value of ZERO when X=+1 and when X=-2.

 

[HallsofIvy]

Homework Statement

Y₁=(x³+3x²-4)/(x²)

Homework Equations

Slant Asymptote at y=x+3. (X-1) and (X+2) appear to be intercepts in the back of the book. How do I factor the numerator to get that?

The Attempt at a Solution

I know this is simple and there is a method to find the zeros of the numerator that I am overlooking- please help.

The only integer factors of 4 are (1)(4) and (2)(2). As symbolipoint said, by the "rational roots theorem" the only rational (in this case, integer) roots must be factors of 4: \pm 1, \pm 2, \pm 4. It easy to try those in the polynomial and see that x= 1 makes it 0: x-1 is a factor. Trying x= -2 also gives a root so x+2 is also a factor.