Finding the limit of (x³-2x²-9)/(x²-2x-3) as x→3

[asd1249jf]

Homework Statement

Find the limit of

$\frac{x^3-2x^2-9}{x^2-2x-3}$

as x->3

Homework Equations

The Attempt at a Solution

You factor the bottom portion and top portion, then it looks something like this

$\frac{x(x^2-2x)-9}{(x-3)(x+1)}$

I feel like I can go further about eliminating the demonimator but I don't know what

 

[dextercioby]

Is 3 a root of the polynomial x^3- 2x^2 - 9 ? If so, what is then factoring of this polynomial ?

 

[Mark44]

(x - 3) is a factor of the numerator. You can use either synthetic division or plain old polynomial division to find the other factor.

 

[SammyS]

... or notice that

$x^3-2x^2-9 = x^3-3x^2+x^2-9$

and factor by grouping.

 

[HallsofIvy]

Homework Statement

Find the limit of

$\frac{x^3-2x^2-9}{x^2-2x-3}$

as x->3

Homework Equations

The Attempt at a Solution

You factor the bottom portion and top portion, then it looks something like this

$\frac{x(x^2-2x)-9}{(x-3)(x+1)}$

No, you did NOT factor the numerator. That is not what "factor" means.

I assume you tried first just putting x= 3 into the fraction and found that both numerator and denominator were 0 when x= 3. The fact that the numerator was 0 tells you that it has a factor of x- 3. $x^3- 2x^2- 9= (x- 3)(ax^2+ bx+ c)$
It shouldn't be hard to see what a, b, and c must be.

I feel like I can go further about eliminating the demonimator but I don't know what