Question

What are the excluded values?
-7
----------
x^2-2x-15
0 and 15
3 and -5
-3 and 5

Answer 1

the excluded values are the values of x that would make the denominator zero in the fraction:\[\frac{-7}{x^2-2x-15}\]so you need to factorise the denominator first and then see which values of x would result in zero for that expression.

Answer 2

so the 1st one

Answer 3

first factorise the denominator - what do you get for this?

Answer 4

how do i do this

Answer 5

you might find this site helpful if you want to learn how to factorise quadratics: http://www.purplemath.com/modules/factquad.htm

Answer 6

so its like (X+2)(x+15)

Answer 7

right?

Answer 8

then its squared

Answer 9

your factorisation is not correct as:\[(x+2)(x+15)=x(x+15)+2(x+15)=x^2+15x+2x+30=x^2+17x+30\]which is not the same as \(x^2-2x-15\).

Answer 10

im confused

Answer 11

i need to know how to do this for a big test im taking

Answer 12

So you are looking for the factors of the expression\[x^2-2x-15\] So what number m and n exist that when we multiply\[(x+m)(x+n)=x^2-2x-15\]

Answer 13

In order for this to work, we know that m+n =-2 and mn=-15. So what two factors -15 can be added to each other to get -2?

Answer 14

which one is m and which one is n

Answer 15

the last one

Answer 16

then

Answer 17

I'm not sure what you mean by "which one" and the "last one" But let's take a guess. If you look at the page that @asnaseer mentioned, some of this will be easier to understand.

Answer 18

im saying i got the last answer

Answer 19

i still dont get with that website

Answer 20

We can guess what m and n to use. Let's say m = -3 and n = 5. Then mn = -15 but m + n = 2. So that isn't right. Let's try m = 3 and n= -5. Now, mn = -15 and m+n = -2. That seems to work. So it looks like\[(x+3)(x-5)=x^2-2x-15\]. We know that the original fraction \[\frac{-7}{x^2-2x-15}\]is undefined if \[x^2-2x-15=0\]

Answer 21

ok

Answer 22

i dont mean to rush but i need to get this question done so i can go to bed

Answer 23

(x+3)(x-5)=0
x=-3, x=5

Answer 24

ok thanks

Answer 25

you're welcome