1. Factoring explanation

This being my first post, let me take the time to say to the people on this board who take the time out of their lives to use their giant brains to help dopes like me, that help is incredibly appreciated.

Now, i need to reduce some fractions and the paper says "it may be necessary to factor before reducing" it then goes on the say that

(x^2) - (xy)
becomes
x(x - y)

and

(x^2) - (y^2)
becomes
(x + y)(x - y)

(with the (x - y)s cancelling out to leave x over (x + y) as the answer)

can someone explain how to do the factoring? i'm clueless.

if someone could do this example, it'd help a bundle too.

(x^2) + (3xy)
--------------
(x^2) - (9y^2)

how would i factor to reduce that?

THANK YOU.

2. Originally Posted by ScottPaulsen1
This being my first post, let me take the time to say to the people on this board who take the time out of their lives to use their giant brains to help dopes like me, that help is incredibly appreciated.

Now, i need to reduce some fractions and the paper says "it may be necessary to factor before reducing" it then goes on the say that

(x^2) - (xy)
becomes
x(x - y)

and

(x^2) - (y^2)
becomes
(x + y)(x - y)

(with the (x - y)s cancelling out to leave x over (x + y) as the answer)

can someone explain how to do the factoring? i'm clueless.

if someone could do this example, it'd help a bundle too.

(x^2) + (3xy)
--------------
(x^2) - (9y^2)

how would i factor to reduce that?

THANK YOU.
You can factor
$\displaystyle a^2 - b^2$
explicitly by using a trick: add and subtract the same number. This does not change the value of the expression, but it does change the form.

$\displaystyle a^2 - b^2 = a^2 + (ab - ab) - b^2 = (a^2 + ab) + (-ab - b^2) = a(a + b) - b(a + b) = (a - b)(a + b)$

However most of us just memorize
$\displaystyle a^2 - b^2 = (a + b)(a - b)$.

$\displaystyle \frac{x^2 - xy}{x^2 - y^2} = \frac{x(x - y)}{(x + y)(x - y)}$

Notice that we now have a common factor we can cancel, the $\displaystyle x - y$. We can cancel this as long as we don't have $\displaystyle x = y$: we require that $\displaystyle x \neq y$.

So finally:
$\displaystyle \frac{x^2 - xy}{x^2 - y^2} = \frac{x(x - y)}{(x + y)(x - y)} = \frac{x}{x + y};~x \neq y$

$\displaystyle \frac{x^2 + 3xy}{x^2 - 9y^2}$

Notice that your denominator is of the form $\displaystyle a^2 - b^2$ where $\displaystyle a = x \text{ and }b = 3y$, so
$\displaystyle x^2 - 9y^2 = x^2 - (3y)^2 = (x + (3y))(x - (3y)) = (x + 3y)(x - 3y)$.

So:
$\displaystyle \frac{x^2 + 3xy}{x^2 - 9y^2} = \frac{x(x + 3y)}{(x + 3y)(x - 3y)}$

Notice that we now have a common factor we can cancel, the $\displaystyle x + 3y$. We can cancel this as long as we don't have $\displaystyle x = -3y$: we require that $\displaystyle x \neq -3y$.

So finally:
$\displaystyle \frac{x^2 + 3xy}{x^2 - 9y^2} = \frac{x(x + 3y)}{(x + 3y)(x - 3y)} = \frac{x}{x - 3y};~x \neq -3y$

-Dan