Question

(5x+3)25x^2-9/5x^2+37x-24
divide

Answer 1

Should there be parentheses around 25x^2-9 ?

Answer 2

\[\frac{(5x+3)(25x^2-9)}{5x^2+37x-24}\]

Answer 3

Is that the problem?

Answer 4

no

Answer 5

simplify your answer

Answer 6

What is the problem then?

Answer 7

reverse the problem

Answer 8

Because this is what you have posted:
\[(5x+3)25x^2-\frac{9}{5x^2}+37x-24\]

Answer 9

That is the problem you have posted. Do you care to correct it?

Answer 10

i guess you put the numerator on the bottom and the denominator on top

Answer 11

Well, no one can help you until you post the problem with which you want help.

Answer 12

then you simplify your answer

Answer 13

Post the correct problem

Answer 14

i did
you have to reverse the problem to where the numerator on bottom and denominator is on top

Answer 15

Ok. Her goes my last guess as to what problem you want help with:

Answer 16

thank you for all your help

Answer 17

\[\frac{5x^2+37x-24}{(5x+3)(25x^2-9)}\]

Answer 18

Is that the problem or not?

Answer 19

yes

Answer 20

Can you factor?

Answer 21

no

Answer 22

Where did you get this problem and why are you expected to simplify it if you have not been taught to factor?

Answer 23

I'm in a distant learning kind of thing taking college classes

Answer 24

that is why i am on open study right now

Answer 25

It's great that distant learning allows you to take those courses, but you're going to suffer if you don't have the necessary background! You might consider watching these videos to plug that hole: https://www.khanacademy.org/search?page_search_query=factoring

Answer 26

To factor the numerator, multiply the 5 and the -24 and then find 2 factors of that number whose sum is -37.

Answer 27

they do not work

Answer 28

What does not work?

Answer 29

the web site u just gave me

Answer 30

Oh. I see someone else is helping you now. Good luck.

Answer 31

@mertsj No, keep working with her, I have to go, just pointing out a source for info on factoring

Answer 32

she left

Answer 33

Well, it's unfortunate that those videos don't play on your computer, because I have to leave as well.

Answer 34

okay

Answer 35

But I'll walk you through factoring the numerator:

\[5x^2+37x−24\]To factor this, multiply the first coefficient with the last one:
\[5*-24 = -120\]Now you need to find two factors of -120 that add to 37.
-1*120
-2*60
-3*40
hey, there's such a pair: -3 + 40 = 37

Now rewrite the middle term as two terms, using those factors as the coefficients:
\[5x^2 +40x - 3x -24\]Group them with parentheses\[(5x^2+40x) -(3x+24)\]Note that the -24 changed to +24 when I moved the - outside the parentheses.

Now, can you factor each of those groups?

Answer 36

ok

Answer 37

what do you get?

Answer 38

do not know

Answer 39

Let's do the left one first:
\[5x^2+40x\]Are there any factors common to both of those terms? If so, what are they?

Answer 40

no

Answer 41

:-(

Can you multiply 5x(x+8) for me?

Answer 42

no don't know how

Answer 43

Okay. This is very basic, beginning algebra. If you don't know this, there's virtually no possibility of success in the class you're seeking help for.

Answer 44

just help me please

Answer 45

Distributive property:

\[a(b+c) = a*b + a*c\]
\[5x(x+8) = 5x*x + 5x*8\]Can you simplify the part on the right?

Answer 46

no

Answer 47

What is your math background?

Answer 48

Well, I'll finish factoring that numerator for you:

\[(5x^2+40x) -(3x+24)\]\[5x(x+8)-3(x+8)\]Now we "undo" the distributive property to give us\[(x+8)(5x-3)\]

Our whole fraction is now
\[\frac{(x+8)(5x-3)}{(5x+3)(25x^2−9)}\]We can factor the denominator:
\[(5x+3)(25x^2-9) = (5x+3)(5x-3)(5x+3)\]because the product term \((25x^2-9)\) is a difference of squares:
\[a^2-b^2 = (a+b)(a-b) = a(a-b) + b(a-b) = a*a - a*b + b*a - b*b \]\[= a^2-b^2\]
Here we have \[25x^2-9\]\[a^2-b^2\]For those to match up, \[25x^2=a^2\]\[9=b^2\]so \[\sqrt{25x^2} = \sqrt{a^2}\]or\[5x=a\]and \[\sqrt{9}=\sqrt{b}\]\[3=b\]Which gives us our factoring of \[25x^2-9=a^2-b^2=(a+b)(a-b) = (5x+3)(5x-3)\]

Our fraction is now\[\frac{(x+8)(5x-3)}{(5x+3)(5x+3)(5x-3)}=\frac{(x+8)\cancel{(5x-3)}}{(5x+3)(5x+3)\cancel{(5x-3)}} = \frac{x+8 } {(5x+3)^2 } \]