Question

Find the zeroes of the following cubic function:
f(x)=x^3-8x-3

Answer 1

Not a nice function to factor.
Your best bet is to graph it and locate the zeroes, where the graph touches the x-axis, or where y-0.

Answer 2

Have you covered rational roots theorem?

Answer 3

|constant term| / |coefficient of x^3| = 3/1 = 3
factors of 3 are: 1 and 3
Possible roots are: -1, 1, -3, 3.

Try each one and see which x value makes x^3 - 8x - 3 zero. Once you find one factor, do synthetic division. The quotient will be a quadratic. Use quadratic formula to find the other two roots.

Answer 4

I'm so confused, should I try and factor it?

Answer 5

Have you been taught rational roots theorem?

Answer 6

It's been a while, but doesn't ring a bell. I think I can try graphing it and then seeing where it touches the x axis? But would it be the entire point as my answer?

Answer 7

Simple and with better understanding :-
factor them.
\[\large \bf x^3-8x-3=(x-3)(x^2+3x+1)\]
So,x=3 is first answer

Answer 8

getting it? @Ruth100

Answer 9

how do you know x = 3?

Answer 10

If you are allowed to use a graphing calculator that will be the easiest. Note down all the x-coordinates where the graph crosses the x-axis and those are the zeros/roots of the function.

Answer 11

yeah that makes sense

Answer 12

factorize it,
\[\large \bf x^3-8x-3=0\]
because we have to find zeroes or you can say roots,so we equate it to zero.

then,after factoring,
\[\large \bf (x-3)(x^2+3x+1)=0\]
so either (x-3)=0 OR \(\large \bf x^2+3x+1\)=0

so our first answer is x-3=0.So, x=3

Answer 13

understood? and if you have any doubt,put x=3 in your equation,if it turn into 0,then x=3 is your 1st zero or root

Answer 14

@Ruth100

Answer 15

@mayankdevnani The question that Ruth had before was HOW did you factor out the (x-3). What were the steps?

Answer 16

oh!!..

Answer 17

its so simple,
@ruth100 ,
first i do trial method,
first put x=1,if they satisfy the equation to 0, then 1 is a root.but its not
then,i put x=2,it also doesn't satisfy and finally,i put x=3,it satisfies.So x=3 is our first root.

then next,i divide the equation by x-3 to get another equation or you can say by factorizing it,i got a new equation and then simply it and get other two new roots.

If we divide,
\[\large \bf \frac{x^3-8x-3}{x-3}=x^2+3x+1\]

Answer 18

that is my method to factorize.

Answer 19

I got y= -3
X= 3
And I can't get the third point. I graphed it.

Answer 20

you can't..its real but you can't graph it.

to find other root 2 roots :- \[\large \bf x^2+x+1=0\]
use quadratic formula

Answer 21

correction :-

Answer 22

I can still use the quadratic formula to find the other roots?

Answer 23

yeah

Answer 24

use quadratic formula to find other 2 roots

Answer 25

Would I use the original given function?

Answer 26

I got 8.35
And -0.35

Answer 27

It seems wrong?

Answer 28

@mayankdevnani

Answer 29

So I did my calculations wrong? I'm really confused