Question

5a+20 divided by a+4

Answer 1

=5(a-4)/(a-4)
**(a-4) reduces, which leaves with : 5

Answer 2

so, it wants the restrictions of the variables so would it be \[a\]

Answer 3

a cannot equal 5?

Answer 4

what value makes the denominator a+4 equal to 0 ?

Answer 5

-4?

Answer 6

so \[\Large a \neq -4\]

to avoid division by zero

Answer 7

Oops I though it was reducing question

Answer 8

that reads "a is not equal to -4"

Answer 9

okay, well.. for the next one it's 2x-3² divided by 5x³ and it says that a cannot equal 0, but how do i solve the problem to get the answer?

Answer 10

same question: what makes the denominator 0?

Answer 11

to answer that, you set the denominator equal to zero and solve for x

Answer 12

so... 1/5?

Answer 13

no

Answer 14

5x^3 = 0

x^3 = 0

x = cube root of 0

x = 0

Answer 15

you can check it by plugging x =0 into 5x^3

Answer 16

that's just 0 though right...?

Answer 17

so that's why x cannot equal 0

to avoid division by zero in "2x-3² divided by 5x³ "

Answer 18

but how do i solve the problem? it's either 1/5, 1/5x, 1/5x², or 2-3x/5x²

Answer 19

oh you just want to simplify

Answer 20

one sec

Answer 21

alright, thanks

Answer 22

The original problem is

\[\Large \frac{2x - 3x^2}{5x^3}\]

right?

Answer 23

yes sir

Answer 24

what can you factor out of the numerator?

Answer 25

ummmmm........ 3x².?

Answer 26

what do 2x and 3x^2 have in common? what factors?

Answer 27

oh, 6?

Answer 28

6 is the LCM of 2 and 3, I'm looking for the GCF

Answer 29

2x has an x

3x^2 = 3*x*x


So they both have 1 x

That's the GCF

Answer 30

when you factor x from the numerator, you get?

Answer 31

1?

Answer 32

2x - 3x^2

x*2 - x*3x

what's the next step? We want to factor out x

Answer 33

divide by 1x?

Answer 34

yes, and have an x outside like this

x(2 - 3x)

Answer 35

if you were to distribute that x back in, you'll get 2x - 3x^2 again

Answer 36

so we factor the numerator and denominator and then cancel to get

\[\Large \frac{2x - 3x^2}{5x^3}\]
\[\Large \frac{x(2 - 3x)}{5x^3}\]
\[\Large \frac{x(2 - 3x)}{x*5x^2}\]
\[\large \frac{\color{red}{\cancel{\color{black}{x}}}(2 - 3x)}{\color{red}{\cancel{\color{black}{x}}}*5x^2}\]
\[\Large \frac{2 - 3x}{5x^2}\]

Answer 37

thanks sooo much!

Answer 38

you're welcome