Question

simplify (3x^2 +3x -36) / (x^2-x-6) make sure to list all restrictions

Answer 1

lol no one looked at this when I answered xP

Answer 2

(3x^2 +3x -36) / (x^2-x-6)
Factor the expression in the numerator first
(3x^2 +3x -36)
= 3 ( x^2 + x - 12)
= 3 ( x^2 +4x -3x - 12)
= 3 [ x(x+4) - 3(x+4)]
= 3 (x+4)(x-3)

Then, factor the denominator.
(x^2-x-6)
= x^2 - 3x + 2x -6
= x(x-3) + 2(x-3)
= (x+2)(x-3)

Since it's a fraction, the denominator should NOT be zero. If it is zero, the fraction becomes undefined.
So, to find its restriction, put denominator = 0
(x+2)(x-3)=0
x+2 = 0 or x-3 = 0
x = -2 or x=3

Therefore, the restrictions are x ≠-2 AND x≠3

Now, back to the question.
\[\frac{(3x^2 +3x -36)}{(x^2-x-6)}\]\[=\frac{ 3 (x+4)(x-3)}{(x+2)(x-3)}\]Cross out the common factor x-3 , you'll get
\[=\frac{ 3 (x+4)}{(x+2)}\] Clear?

Answer 3

sooo youre supposed to put the denominator equal to zero?

Answer 4

Put the denominator equal to zero to get the restriction - Yes

Answer 5

oh see this makes sense thank you :)

Answer 6

Welcome~~

Answer 7

so you should always find the restrictions right after you factor and before you completely simplify?

Answer 8

Yes, since you can't cancel 0 as a common factor.

Answer 9

okay omg thank you so much! Im gonna have another question up soon ;)

Answer 10

Sometimes it really miff me off that the site doesn't have a edit button :/

\[ \frac {(3x^2 +3x -36)}{ (x^2-x-6)} = \frac{3 (x-3)(x+4)}{(x-3)(x+2)} \]

Now about the restriction, cancelling is a confusing word, the better word is divide both of the numerator and denominator by their gcd. In this case it is (x-3), but as we know that division by zero is undefined, hence \((x-3) \neq 0 \implies x \neq 3\). So we get:\[\frac{3 (x+4)}{(x+2)} = \frac{3x+12}{(x+2)}\]\]

Answer 11

I think it's better to write the final answer as \[ \frac {(3x^2 +3x -36)}{ (x^2-x-6)} =\frac{3x+12}{(x+2)}, \forall x \neq -2 \] after writing all of what I stated before.