Some Math help (factoring)

I need help with these problems... :/

The limit as x approaches -3 of (t^2-9)/(2x^2+7x+3) (I mostly need help with factoring the bottom...)

The limit as x approaches -2 of (x+2)/(x^3+8) (once again, factoring the bottom)

Basically, all I need to know is how to factor the bottom of those two. Please show a step-by-step, it's been ages since I've done factoring. Thanks

Think of most quadratic equations as fitting the format ax^2 + bx + c.

To factor a quadratic equation, do these 2 steps:
1. Take the product of a and c, giving you ac.
2. Find two numbers that multiply to give you ac and that add to give you b.

For 2x^2 + 7x + 3, a = 2, b = 7, and c = 3.
1. Product of 2 and 3: 6.
2. Two numbers that multiply to give you 6 and add to give you 7: 6 and 1.

Now change 2x^2 + 7x + 3 into 2x^2 + 6x + 1x + 3.
You can further factor this: 2x(x + 3)+1(x + 3)
Since (x + 3) is common, factor it out: (x + 3)(2x + 1).

So that's how you factor the bottom of the first question, giving you (x + 3)(2x + 1).

For x^3 + 8, since x is cubed, let's change 8 to be a cube. 8 = 2^3 so we can change x3 + 8 into

x^3 + 2^3

= (x + 2)[x^2 - (x)(2) + (2)(2)]

= (x + 2) (x^2 - 2x + 4) (If you combine the two terms, you find that it does equal x^3 + 8.)

So that's the bottom of the second question: (x + 2)(x^2 - 2x + 4).

To factor the bottom of the first rational equation:

(denominator) --> 2x^2+7x+3 factors out to be (2x+1)(x+3)

As for the second rational equation:

x^3+8 is the same thing as x^3+2^3. So, you can say that it is equal to (x+2)^3

lim (t²-9)/(2x²+7x+3) =lim [(t+3)(t-3)]/[((2x+1)(x+3)]
x→-3 x→-3


lim (x+2)/(x³+8) =lim (x+2)/[(x+2)(x²-2x+4)
x→-2 x→-2
=lim(1/(x²-2x+4)
x→-2
=1/(4+4+4)
=1/12

2x^2 +7x+3 =(2x^2+6x)+(x+3)
=2x(x+3)+1(x+3)
= (2x+1)(x+3) ...............Ans
x^3+8 = (x)^3 +(2)^3
= (x+2)(x^2 -2x+4) ...............Ans