Factoring

**Alan306090** · February 15th 2010, 05:23 PM

(2X^2 - 7X - 15)/(2X^3 + 13X^2 + 15X)

If X is greater than 10, then why is (x-5)\(x^2 + 5x) equivalent to it?

And why can't X = -3\2?

**Prove It** · February 15th 2010, 05:36 PM

Originally Posted by Alan306090

(2X^2 - 7X - 15)/(2X^3 + 13X^2 + 15X)

If X is greater than 10, then why is (x-5)\(x^2 + 5x) equivalent to it?

And why can't X = -3\2?

$\frac{2x^2 - 7x - 15}{2x^3 + 13x^2 + 15x}$

$= \frac{2x^2 - 10x + 3x - 15}{x(2x^2 + 13x + 15)}$

$= \frac{2x(x - 5) + 3(x - 5)}{x(2x^2 + 10x + 3x + 15)}$

$= \frac{(x - 5)(2x + 3)}{x[2x(x + 5) + 3(x + 5)]}$

$= \frac{(x - 5)(2x + 3)}{x(x + 5)(2x + 3)}$

Can you see here that you can't let $x = -\frac{3}{2}$ ? If you did the denominator would be $0$ .

Now cancelling gives

$= \frac{x - 5}{x(x + 5)}$

$= \frac{x - 5}{x^2 + 5x}$ .

**Archie Meade** · February 15th 2010, 05:45 PM

Originally Posted by Alan306090

(2X^2 - 7X - 15)/(2X^3 + 13X^2 + 15X)

If X is greater than 10, then why is (x-5)\(x^2 + 5x) equivalent to it?

And why can't X = -3\2?

$\frac{2x^2-7x-15}{2x^3+13x^2+15x}=\frac{(2x+3)(x-5)}{x(2x+3)(x+5)}$

This has no value if x=0, x= -5 and 2x = -3.

You can cancel the term (2x+3) only if (2x+3) is not zero.
Since in that case you have 0/0.

However, if x>0, you can cancel those terms and then we can say it's the same as

$\frac{x-5}{x(x+5)}$

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