# Thread: Lots of confusion on factoring this...

1. ## Lots of confusion on factoring this...

The function f is given by the first formula
when x < -2 and

by the second formula
when -2 is less than or equal to x.

I can't figure out what value I have to put in for a to make the function continuous at -2?

2. Originally Posted by frozenflames

The function f is given by the first formula
when x < -2 and

by the second formula
when -2 is less than or equal to x.

I can't figure out what value I have to put in for a to make the function continuous at -2?
The only problem with finding the limit- and the only reason why you want to factor- is that both numerator and denominator are 0 when x= -12. And that tells you that (x- (-2))= x+ 2 is a factor. Divide $4x^3+ 8x^2+ 8x+ 16$ by x+ 2 ("synthetic division" works nicely here) to find the other factor.

3. Originally Posted by frozenflames

The function f is given by the first formula
when x < -2 and

by the second formula
when -2 is less than or equal to x. x > -2

I can't figure out what value I have to put in for a to make the function continuous at -2?
note ...

$\frac{4x^3+8x^2+8x+16}{x+2} = \frac{4x^2(x+2) + 8(x+2)}{x+2} = 4x^2+8$

to be continuous at $x = -2$, $\lim_{x \to -2} f(x)$ must exist ...

$\lim_{x \to -2^+} x^2 - 6x + a = \lim_{x \to -2^-} 4x^2 + 8$

$a =$ ?

4. 8!

That makes sense. The reason why I was flustered was because I forgot the x^2 and converted the first equation to [(4x+8)(x+2)]/(x+2).

Thanks!