# Lowest terms

• Apr 7th 2009, 10:26 PM
Tenskypoo
Lowest terms
Multiply or divide, write each answer in lowest terms.

3x^3-5x-2
------------
x-2

times

x-3
----
x+1

Where do i begin?

(corrected, format looked odd before)
• Apr 7th 2009, 10:52 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
Multiply or divide, write each answer in lowest terms.

3x^3-5x-2
------------
x-2

times

x-3
----
x+1

Where do i begin?

begin by factoring $\displaystyle 3x^2 - 5x - 2$ and see what cancels. i suppose the power of the first x is a 2 and not 3 based on the level of questions you have asked before.
• Apr 7th 2009, 11:02 PM
Tenskypoo
after doing

$\displaystyle 3x^2 - 5x - 2$

i came up with

$\displaystyle ( 3x + 1) ( x - 2)$

which eliminates the x-2 from the bottom of the other problem, leaving me with 3x+1?
• Apr 7th 2009, 11:20 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
after doing

$\displaystyle 3x^2 - 5x - 2$

i came up with

$\displaystyle ( 3x + 1) ( x - 2)$

which eliminates the x-2 from the bottom of the other problem, leaving me with 3x+1?

• Apr 7th 2009, 11:25 PM
Tenskypoo
well i have to multiply $\displaystyle 3x + 1$

by x-3 over x+1, how do i do that?

i want to GUESS. a heavy guess. that you rule out x+1 and you're left with x-3 over 3, or vise versa.
• Apr 7th 2009, 11:44 PM
Jhevon
Quote:

Originally Posted by Tenskypoo
well i have to multiply $\displaystyle 3x + 1$

by x-3 over x+1, how do i do that?

i want to GUESS. a heavy guess. that you rule out x+1 and you're left with x-3 over 3, or vise versa.

that would be wrong. you cannot rule out the x + 1 from the 3x + 1. if it were 3(x + 1) you could. but that is not the case. just multiply the numerators to get the new numerator, and the denominators to get the new denominator
• Apr 7th 2009, 11:50 PM
Tenskypoo
I don't want to seem like a complete idiot, but could you possibly show me what the final set up before multiplying should be?
• Apr 8th 2009, 12:16 AM
Jhevon
$\displaystyle \frac {3x^2 - 5x - 2}{x - 2} \times \frac {x - 3}{x + 1} = \frac {(3x + 1)(x - 2)}{(x - 2)} \times \frac {x - 3}{x + 1} = (3x + 1) \times \frac {x - 3}{x + 1} = \frac {(3x + 1)(x - 3)}{x + 1}$