Question

find the given zeros of the function

f(x)=x^3+3x^2-6x-18

f(x)=x^3+3x^2+6x-18

f(x)=x^3+3x^2-6x+18

Answer 1

factor first two terms and
factor last two terms

Answer 2

i know, i do but i get like plus or minus negative 6 so do i pull the negative out or?

Answer 3

f(x)=x^3+3x^2-6x-18
------- ------

= x^2(x+3) - 6(x+3)

Answer 4

^^yess

Answer 5

next factor the gcf again

Answer 6

f(x)=x^3+3x^2-6x-18
------- ------

= x^2(x+3) - 6(x+3)

= (x+3)(x^2-6)

Answer 7

you need to factor x^2-6 still

Answer 8

u may use the identity : a^2-b^2 = (a+b)(a-b)

Answer 9

so @ganeshie8 which one would give me the the given zeroes of the square root of 6, the negative square root of 6, and -3???

Answer 10

x^2-6 = x^2 - sqrt(6)^2 = (x+sqrt(6))(x-sqrt(6))

Answer 11

plug that in,

f(x)=x^3+3x^2-6x-18
------- ------

= x^2(x+3) - 6(x+3)

= (x+3)(x^2-6)

= (x+3) (x+sqrt(6))(x-sqrt(6))

Answer 12

thank you!