Question

PLEASEEE HELP!!!
What is the domain of (√x+8)/(x+4)(x-6)
x ≠ -8, x ≠ -4, x ≠ 6

x > 0

All real numbers

x ≥ -8, x ≠ -4, x ≠ 6

Answer 1

@KlOwNlOvE @tkhunny @SolomonZelman

Answer 2

square roots need to have values which are >= 0 and denominator cannot have any vaule which make them 0

Answer 3

so x+8>=0 and x is not -4 and x is not 6

Answer 4

the last one

Answer 5

I love you a lot! lol thank you

Answer 6

Do you think you can help me with one more?

Answer 7

ur welcome:) sure, if i can

Answer 8

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.
5, -3, and -1 + 3i

Answer 9

(x) = x4 + 12.5x2 - 50x - 150

f(x) = x4 - 4x3 + 15x2 + 25x + 150

f(x) = x4 - 4x3 - 15x2 - 25x - 150

f(x) = x4 - 9x2 - 50x - 150

Answer 10

do you know about complex numbers ?

Answer 11

they are of the form a + bi, and i^2=-1 you should know it

Answer 12

and it is a rule that if a polynomial has a 0 which is a+bi, then the other 0 should be a-bi, then here we have -1+3i so the other 0 should be -1+3i, and others are 5 and -3 as given

Answer 13

lets add them: -5 + (-3) + (-1+3i) + (-1-3i) = -10

Answer 14

so the sum of 0s are -10

Answer 15

What do I do with the -10?

Answer 16

yeah i was looking into it, ok so we can construct an equation in the form (x+a)(x+b)(x+c)(x+d)

Answer 17

lets try it: we have all the zeros so it will be like: (x+5)(x+3)(x+1-3i)(x+1+3i)

Answer 18

we should now multiply them.. oops by the way the form is (x-a)(x-b)(x-c)(x-d)=0 if the 0s are a,b,c and d

Answer 19

so lets first multiply the first two terms: (x+5)(x+3)=x^2+8x+15

Answer 20

Wouldn't they be negative though?

Answer 21

then lets multiplty the last two: ((x+1)-3i)((x+1)+3i) = ((x+1)^2)+3i(x+1) - 3i(x+1) -9i^2

Answer 22

= x^2+1+2x + 9

Answer 23

so we can now multiply the first result with this: (x^2+8x+15)(x^2+2x+10) = x^4+2x^3+10x^2 + 8x^3+16x^2+80x + 15x^2+30x+30 = x^4+10x^3+26x^2+110x+30

Answer 24

sth is wrong...

Answer 25

yeaaaa it should be x-5 above, i worte x+5

Answer 26

That isn't one of my options though

Answer 27

x) = x4 + 12.5x2 - 50x - 150

f(x) = x4 - 4x3 + 15x2 + 25x + 150

f(x) = x4 - 4x3 - 15x2 - 25x - 150

f(x) = x4 - 9x2 - 50x - 150

Answer 28

let me revise: (x-5)(x+3) = x^2-2x-15

Answer 29

so (x^2-2x-15)(x^2+2x+10) = (x^4+2x^3+10x^2 - 2x^3 -4x^2-20x - 15x^2-30x-150 = x^4 -9x^2-50x-150

Answer 30

the last one

Answer 31

Thank you for all your help. Youre the best :))

Answer 32

np, but please notice what we did: we first thought that if a polynomial has a complex 0 in it in the form a+bi, then the other 0 should be a-bi; and if we multiply all the 0s then we can get the equation.. know that the equation is of the form (x-a)(x-b)(x-c)(x-d)=0 where a, b, c, and d are all 0s (the 3 of which are given in the question and the last one we found as -1-3i) and we multiplied them all.. Notice how i multiplied the complex numbers: i grouped the real parts and imaginary parts together and multiplied them that way and also know that i^2 = -1

Answer 33

I understand :) @barbillus