Question

(2X^2-98)/(X^2+4X-21)=

Answer 1

\[\frac{2x^{2}-98}{x^{2}+4x-21} = \frac{2(x^{2}-49)}{(x+7)(x-3)} = \frac{2(x+7)(x-7)}{(x+7)(x-3)} = \frac{2(x-7)}{x-3}\]

Answer 2

Basically just factor everything and things will cancel

Answer 3

2 factors out of each term on top.
Which gives us a numerator of:
2(x^2-49)

Now for the denominator:
We know that 7*3 = 21, but we need to find a combination of 7 and 3 that will make 4 when added together.
7+3 = 10
-7 + 3 = -4
-7 - 3 = -10
7 - 3 = 4

So we are looking for x + 7 and x - 3
So our denominator factors to (x+7)(x-3)

We have the new equation of 2(x^2-49)/(x+7)(x-3)

Now we are looking for a factored equation which will become x^2-49.
We know that 7 * 7 is 49.
If we add 7 and -7 we have 0, which is the inner term above. (x^2+0x-49)
So we now have:

2(x+7)(x-7)/(x+7)(x-3)

Now the x+7 above and below will cancel.
2(x-7)/(x-3)

Hope this helps.
http://www.tutorsean.net