Question

How to find the zeroes of a function?
y = x^3 – 7x^2 – 18x

Answer 1

@mathstudent55

Answer 2

or @ganeshie8 Could either of you help me?

Answer 3

ok I just figured this out a lil bit so lets try.

Answer 4

set it equal to zero, and factor out of x.

Answer 5

y= f(x) = x^3 – 7x^2 – 18x = 0
x(x^2 – 7x – 18) = 0
x = 0 (first root)
x^2 - 7x - 18 = 0
put this one in quadratic formula to get other roots :D

Answer 6

\(\LARGE\color{blue}{ \bf y = x^3 – 7x^2 – 18x }\)

\(\LARGE\color{blue}{ \bf x^3 – 7x^2 – 18x =0 }\)
\(\LARGE\color{blue}{ \bf x(x^2 – 7x – 18) =0 ~~~~~~~~~~~~x=0}\)
\(\LARGE\color{blue}{ \bf x^2 – 7x – 18 =0 }\)
\(\LARGE\color{blue}{ \bf (x+2)(x-9) =0 ~~~~~~~x=2,-9}\)

Answer 7

Kitty solved you the first part let me know if you have problem solving the other part

Answer 8

I mean -2 and 9

Answer 9

So, 0, -2, 9

Answer 10

Thanks both of you! I could have gotten the rest myself @SolomonZelman but thank you!

Answer 11

Welcome :D I know right I'm a cute kitty :3 *roof roof* lol

Answer 12

RoseDryer , not a problem!

Answer 13

So for this one
It would be
y = x(x + 6)
x(x+6)=0
x=0
x+6=0
x=-6
x=0,-6?

Answer 14

Yes, you are indeed correct.

Answer 15

Great!

Answer 16

Yes :)

Answer 17

y = (x + 1)(x – 1)(x – 2)
Do I just set each one equal to 0?

Answer 18

YES basically, b/c if
(x + 1)(x – 1)(x – 2)=0
then one of these
(x+1) , (x-1) or (x-2) is zero.

Answer 19

Okay

Answer 20

For this one do I divide by 3
y = 3x3 – 3x
and get
y=x^3-x?

Answer 21

try 3x (x^2-1)=0
3x=0 , x^2-1=0
x=0 , x^2 + 0x - 1=0
apply quadratic formula and get other two roots

Answer 22

\[y = 3x^{3} – 3x \]\[ 3x^{3} – 3x =0\]\[ 3(x^{3} –x )=0\]\[x^{3} –x =0\]\[x^{3} =x\]\[x=~~±~1,0\]