Question

Show all work in simplifying the quantity of three x squared minus three x minus sixty, divided by the quantity of x squared plus nine x plus twenty Be sure to list restrictions.

Answer 1

\[3x^2-3x-60 \over x^2+9x+20\]

Answer 2

factorise both the top nd the bottom differently nd they shouls cancel out

Answer 3

that's what I am working on so far I have \[(3x+ )(x-) \over (x+4)(x+5)\]

Answer 4

write down the factors of 60

Answer 5

1*60, 2*30, 3*20, 4*15, 5*12, & 6*10

Answer 6

have you tried inputting all the factors?? if it doesnt work devide across the top half only by 3 that gives u X squared -minus X minus 20 which will be much easier to work with

Answer 7

answer for top is (x+4)(x-5)

Answer 8

with your factorisation of the numerator take out 3 as a common factor 1st

\[\frac{3(x^2 - x - 20)}{x^2 + 9x + 20}\]

now factor the quadratics.

Answer 9

@campbell would it be the same as deviding across top half by 3?

Answer 10

no.... because of you are simplifying an expression....

if you divide the numerator by 3 you would also have to divide the denominator by 3


the 3 will play a roll in the final answer

Answer 11

thank you very much for explaining that to me im a bit rusty with my maths since the summer holidays :)

Answer 12

ok, so then what happens after it becomes \[3(x^2-x-20) \over (x+4)(x+5)\] ?

Answer 13

for the numerator find the factors of -20 that add to -5, the larger factor is negative and the smaller is positive.

Answer 14

that will be 4 and 5

Answer 15

which is negative and which is positive..?

Answer 16

5 is negative 4 is positive

Answer 17

ok so now it looks like

\[\frac{3(x +4)(x -5)}{(x +4)(x +5)}\]

so next you need to look at restrictions before simplifying

Answer 18

if I am not mistaken as they r multiplied by each other the (x+4) at the top and bottom should cancel each other

Answer 19

they do cancel but there is a restriction.... a point of discontinuity at x = -4

and there is a vertical asymptote at the other restiction.

Answer 20

is one of the restrictions \[\neq-5\] ?

Answer 21

thats correct... well done

Answer 22

is there only one restriction?

Answer 23

there is a horizontal restriction.... y = 3

Answer 24

hmm I dont understand this restrictions stuff nd asymptotes btw what level maths is this??

Answer 25

algebra 2

Answer 26

hmm not sure im familiar with that hope u dont mind me tagging along and learning too from campbell

Answer 27

so now you cancel out the x+4 ...... \[3(x-5) \over (x-5) \] with \[\neq-5 \] and \[\neq3\] now what do you do?

Answer 28

after cancelling you get

\[\frac{3(x -5)}{x + 5}\]

Answer 29

oops sorry i mixed up the signs for the bottom one

Answer 30

so the restictions.... vertical asymptotes

\[x \ne -4, -5\]

horizontal asymptote

\[y \ne 3\]