Question

4x^3-17x^2-42x/x^2y-13xy+42y simplify

Answer 1

\[\large \frac{4x^3-17x^2-42x}{x^2y-13xy+42y}\]

Answer 2

yes, now it needs to be simplified
=]

Answer 3

I just needed to see it written in equation format first lol. One min. :P

Answer 4

thank you!

Answer 5

4x^3-17x^2-42x so in case of this what can you factorizing out ?

Answer 6

All i've got so far is that..

You can factor out an x from the numerator, and a y from the denominator.

\[\large \frac{x(4x^2-17x-42)}{y(x^2-13x+42)}\]

Answer 7

Now can you not solve them separately? The numerator and denominator?

Answer 8

well. i dont know lol. Its mulitple choice but none of it is even close =/ other than the x and y being factored out

Answer 9

We an factor out the bottom though :) so that's a plus.

What two numbers add to give -13 and multiply to give +42?

Answer 10

-7 -6

Answer 11

Good. \[\large \frac{x(4x^2-17x-42)}{y(x-7)(x-6)}\] So far...

Answer 12

ok

and the numerator ?

Answer 13

factor out

Answer 14

first we can split the 4x^(2) into

(x )(4x )

Then analyzing the factors of 42, we understand that +7 and -6 = -42, and adding them together gives you 13, BUT we have a 4 present also, so 13+4 = 17.
Knowing this, we an simplify 4x^(2) -17x -42 into \[\large (x-6)(4x+7)\]

Answer 15

does the x-6 on top and bottom cancel

Answer 16

Yep it cancels out.\[\large \frac{x(4x^2-17x-42)}{y(x-7)(x-6)}= \frac{x\cancel{(x-6)}(4x+7)}{y(x-7)\cancel{(x-6)}}\] and what do you get?

Answer 17

do you bring the 4x outside?

Answer 18

Nope.

Answer 19

nvm lol

Answer 20

thank you!

Answer 21

You can simplify the numerator and denominator by distributing the x and y back into their respective places.

Answer 22

and no problem :)

Answer 23

are you good at restrictions? lol

Answer 24

Depends, haha.,

Answer 25

f(x)= x+1/ x^2+5x-6

Answer 26

gotta find the restriction

Answer 27

How do you simplify x^(2) +5x - 6?

What two numbers add together to give 5, but multiply to give -6

Answer 28

-1 6

Answer 29

so is that the restriction? -1,6

Answer 30

Alright, so you have \[\large \frac{x+1}{(x+6)(x-1)}\] solving for the denominator, set both = 0
\[\large x+6=0\]\[\large x-1=0\] and solving for these x values will give you yourrestriction.

Answer 31

so x= -6 x=1

Answer 32

mmhmm :)

Answer 33

and to double check this, plug these values back into your original function \[\large f(x) = \frac{x+1}{x^2+5x-6}\]\[\large f(1)=\frac{2}{0} = \text{undefined}\]\[\large f(-6)= \frac{-6+1}{(-6)^2+5(-6)-6}=\frac{-5}{0}= \text{undefined}\]

Answer 34

thanks!

Answer 35

np :)

Answer 36

i dont know how you write the problems our on here, can you write this out for me?

Answer 37

x+2/7-x * x^2-9x+4/x^2-5x+6

Answer 38

the x+2 is divided by the 7-x

Answer 39

and same for the other side

Answer 40

Use the equation editor :P