Question

Pretest for school. No grade just need a better understanding as well as a solution? I want to be prepared and have a good piece of mind walking into school monday. "Simplify by algebraic ratios: x^4 - 3x^3 over x^2 - 9

Answer 1

hint-
x^2-9 = x^2-3^2

( you can factor this using the identity a^2-b^2 = (a+b)(a-b) )

Answer 2

does that help.... you can finish the problem wid that hint ?

Answer 3

I still can't do the bottom one. I have never learned this before. It's a semester pretest.

Answer 4

ohhk... its easy. bottom can be written like this :

x^2-9 = x^2-3^2 = (x+3)(x-3)

Answer 5

Oh :D thanks that helped a lot

Answer 6

glad to hear that ! (: top you can take x^3 common, right ?

Answer 7

yes so it's (x-3)(x+3) over (x+3)(x-3)

Answer 8

\(\huge \frac{x^4-3x^3}{(x+3)(x-3)}\)

Answer 9

Oh

Answer 10

I was still off haha

Answer 11

thats the given expression.... we just factored the bottom part.

Answer 12

lets factor top... okay ?

Answer 13

and see if something cancels

Answer 14

\(\huge \frac{x^4-3x^3}{(x+3)(x-3)}\)

\(\huge \frac{x^3(x-3)}{(x+3)(x-3)}\)

\(\huge \frac{x^3\cancel{(x-3)}}{(x+3)\cancel{(x-3)}}\)


\(\huge \frac{x^3}{(x+3)}\)

Answer 15

Thank you!

Answer 16

complete solution there. see if it makes sense....

Answer 17

Way better understanding of it now.

Answer 18

great ! yw.. .. . :D