Question

please help!!!! x^2/x^2+3*x+14/x...............multiply and simplify to lowest terms

Answer 1

notice how there is an x^2 term in the first numerator and an 'x' term in the second denominator

Answer 2

so x^2 divided by x = x

Answer 3

which means that

\[\Large \frac{x^2}{x^2+3}*\frac{x+14}{x}\]

becomes

\[\Large \frac{x}{x^2+3}*\frac{x+14}{1}\]

Answer 4

From here, you multiply the numerators together to get a final numerator

then you multiply the denominators together to get a final denominator

Answer 5

@jim_thompson5910, thanks for your help, so i came up with 14x^2/4x^2???

Answer 6

hmm that's not correct

Answer 7

or should it have been 14x^2/3x^3

Answer 8

You should get this

\[\Large \frac{x^2}{x^2+3}*\frac{x+14}{x}\]

\[\Large \frac{x}{x^2+3}*\frac{x+14}{1}\]

\[\Large \frac{x(x+14)}{(x^2+3)*1}\]

\[\Large \frac{x^2+14x}{x^2+3}\]

Answer 9

oh ok I see, what I done wrong

Answer 10

glad you do

Answer 11

thanks... is division pretty much the same way??

Answer 12

with division, you flip the second fraction and you multiply

Answer 13

ok can you just look at this one and tell me if Im way off or not??

Answer 14

sure go ahead and post it

Answer 15

5x^2-15x/2x-9 divided by x-3/2x-9...sorry could get the equations to work but I came up with 5x^2-15/x-3

Answer 16

you should have 5x^2-15x/x-3

Then you factor the numerator:

5x^2 - 15x = 5x(x-3)

The 'x-3' terms will cancel leaving you with just 5x as the final answer

Answer 17

okay I get it..thank you again

Answer 18

yw