Question

Determine the zeros of f(x) = x^3 – 3x^2 – 16x + 48

Answer 1

can you please go one step at a time

Answer 2

Determine the zeros of f(x) = x^3 – 3x^2 – 16x + 48

The question asks you find the values of x for which f(x) is 0. Hence @callisto set \(f(x)=0 \)

Answer 3

yeah!
\[f(x)=x^3-3x^2-16x+48=0.\]

Answer 4

The next step is taking common factors from first 2 & last 2.

Answer 5

After that, the method employed is known as factorization of a cubic polynomial. It's basically just to factor out the common things and reduce it to a quadratic and then factor it to finally reduce it to three linear equation in product form.

Finally, it's equating every term to zero and reducing the final answer :)

Answer 6

Yeah: ^_^

Answer 7

To understand the last step

Think about, \(a\times b = 0\), this can only happen when either of them is zero.

Answer 8

The same is true about \( a\times b\times c =0 \)

Answer 9

@Callisto Why u deleted the solution?
We were trying to make her understand ur solution:)

Answer 10

ur solution was best:)

Answer 11

It was not good at all....

Answer 12

Hm but without your solution, our explanation are somewhat meaningless.

Answer 13

hmm........
i don't agree.
becoz the second method is too hard for @sara1234 to understand.

Answer 14

so ur sol. was best @ all.

Answer 15

okay so can we start by doing the rational root thrm

Answer 16

Yes we can use rational root theorem. But it is a bit tedious compared to the solution posted by @callisto

Answer 17

Put f(x) =0
x^3 – 3x^2 – 16x + 48 = 0
x^2 (x-3) -16 ( x-3) = 0
(x^2-16) (x-3) =0
(x-4)(x+4)(x-3) =0
x+4 =0 or x-3 =0 or x-4=0
x = -4 or x =3 or x=4

Answer 18

can we do it the other way I think it's easier for me to understand

Answer 19

Factors of 48- +-1=+-2+-3+-4+-6+-8+-12+-16+-24+-48
Factors of x- +-1

Answer 20

We can find roots according to this too:)

\[ax^3+bx^2+cx+d=x^3+(\alpha+\beta+\gamma)x+(\alpha \beta+\beta \gamma+\alpha \gamma)x+\alpha \beta \gamma.\]Where \[\color{blue}{\alpha, \beta, \gamma \space \text{are roots of this cubic eq..}}\]

Answer 21

Using the rational root theorem, the possible roots are of the form:

\[ \pm \frac{1, 2, 3, 4, 6, 8, 12, 16, 24, 48}{1} \]

If you go through each of them you will see that that equation will yield zero at \(x=3,\pm4 \)

Answer 22

okay so now we have to do rule of signs

Answer 23

2 or 0 positive real zero

Answer 24

right?

Answer 25

From my solution
if we break \[\alpha \beta \gamma=+48.\]Such that \[\alpha+\beta+\gamma=-3.\]then we get \[\alpha=+4;\beta=-4;\gamma=-3\]So our roots will be
\[- \alpha; -\beta ; -\gamma\]

Answer 26

@maheshmeghwal9 i really don't understand what your trying to explain to me

Answer 27

@maheshmeghwal9 's approach is also known as vieta's theorem.

Answer 28

http://mathworld.wolfram.com/VietasTheorem.html

Answer 29

yeah of course:)
I know that @im2bad
no need to understand @sara1234 I was only showing another method

Answer 30

just concentrate on @Callisto 's method :D

Answer 31

sorry im not that smart as you guys so can you dumb it up a bit lol

Answer 32

Have you learnt factorization ?

Answer 33

the way i am trying to do this is rational thrm then rule of signs then fundamental thrm then quadratic formula but im having trouble

Answer 34

Hmmm you haven't answered my question :|

Answer 35

factorize & get ur zeroes

Answer 36

@sara1234 did you understand ?

Answer 37

Umm.... i'll just try another question

Answer 38

No, I'll help you. Don't give up

Answer 39

We'll use the rational factor theorem here, if there are any real roots, they will be of the forms
\[\frac{\pm \ factors\ of\ 48}{1}=\frac{\pm 2. 2. 2.2.2 .3}{1}\]
Do you get this?

Answer 40

yes that's what i was trying to say we start of with

Answer 41

Yeah, try +2 or -2, see if any of these two is a zero

Answer 42

2 or 0 positive real zero

Answer 43

That's correct, I meant that check if +2 or -2 is a zero of f(x)
Check if f(2)=0
or f(-2)=0

Answer 44

can we continue this tommorw i have to go i really apperciate your help

Answer 45

Sure thing, tag me to call me :)