Factoring/ Reducing Fractions?

Question

I have a ditto for hw and i cant figure out 2 answers.

1. x3-x2-6x (the 3 and 2 represt cubed and squared as in xcubed and xsquared)
(FACTOR COMPLETELY)

2. {(x2-9)/(2x2+5x-3)}/{(1)/(2x-1)} (if a number appears directly after an x it is a superscript meaning squared, cubed, etc.)
(EXPRESS IN SIMPLEST FORM)
x2-9 1
--------------- / ------------
2x2+5x-3 2x-1

Answer 1

Ok to solve the 1st one....x^3 - x^2 - 6x

1st the best way to solve it is go for the obvious. First find the common factor, which number or variable is in each thing. This is x. So factor this one out. Therefore this leaves

>> x(x^2-x-6)

So if u use the distributive property, you'll get the same problem you started out with.

2nd Problem

>>1st factor the first two. x^2-9 would be (x-3)(x+3)
then 2x^2+5x-3 would be (2x-1)(x+3).
Ok when these go over each other it the (x+3) would cancel each other out. thus leaving (x-3) / (2x -1 ). Ok the bottom one leaves the same.... Then u need to find the lowest common denomonator (lcd) which will be (2x-1) multiple it to both the bottom and top.
Thus leaving...
>>
{( 2x - 1 ) ( x - 3 ) }
------------------------
( 2x - 1 )
---------------------------------------
(2x-1)
----------------------
(2x-1)
Ok you need to cancel out the 2x -1 on the top and bottom. Leaving (x -3 )

this is the answer (x-3)

Well hopefully I didn't lose you....and hopefully the answers actually right....but i'm pretty sure it is....well again hopefully I helped!!!

Answer 2

On the first one, factor out an x first so you have
x(x2 - x - 6) and then it's easy to factor the parentheses:
x (x-3) (x+2)

Answer 3

for the reason that they the two bring about 5, i could immediately p.c.. 5 for area of the two. as a result, a hundred and five divides into 21 and 5, and a hundred sixty five divides into 33 and 5. The 5s divide out, so your very final fraction is 21/33