Question

What are the zeros of the polynomial function f(x) = x3 – 4x2 – 12x?

Answer 1

to find zeros of a polynomial we want to factor to get values of x that make the whole thing 0. \[f(x) = x^3 -4x^2 -12x \] what term is in each part?

Answer 2

a Polynomial of degree 'n' will have n zeros, this is a polynomial of degree 3, therefore there will be 3 values of x that make the whole equation 0.

Answer 3

0, –2, 6
–2, 6
0, 2, –6
2, –6
here is the answers. & I'm still confused.

Answer 4

I understand the (X) is the term in each part.

Answer 5

ok so 'pull out' the 'x'

Answer 6

\[f(x) = x^3 -4x^2-12x = x(x^2 -4x -12)\]

Answer 7

which all equals 0

Answer 8

to find zeros you set f(x) = 0

Answer 9

So its a?

Answer 10

lets keep going until you know for sure.

Answer 11

Yea, cause I am really lost. After you pulled the "x" out i got lost.

Answer 12

do you get how we can pull out the x?

Answer 13

no.

Answer 14

we are reversing the distribution process. \[x(a+b) = ax+bx\]

Answer 15

Okay

Answer 16

here we are starting with the right hand side and going 'backwards'

Answer 17

yea

Answer 18

what value of x makes the following true: x*(anything) = 0

Answer 19

0?

Answer 20

correct!

Answer 21

okay

Answer 22

\[f(x) = 0 = x(x^2-4x-12)\]

Answer 23

so in this problem either 'x' = 0 or \[x^2 - 4x -12 = 0\]

Answer 24

Okay so then how do you get one of these answers?
0, –2, 6
–2, 6
0, 2, –6
2, –6

Answer 25

to find zeroes of a polynomial what we do is take the polynomial and factor it into seperate pieces of x such as " x(x+a)(x+b) = 0 and then to find what values of x makes this true either of the followings need to occur, if x=0 , if (x+a)=0, or if (x+b) = 0

Answer 26

the number of zeroes in a polynomial is equal to the number of the power on the highest exponent, in this case 3

Answer 27

so we are looking for 3 components that any part can equal 0 to make the whole function = 0

Answer 28

So b and d are out.

Answer 29

yes, for this one.

Answer 30

the terms of x(x+a)(x+b) are called 'roots'

Answer 31

Okay, then what do we do.

Answer 32

they correspond to where the function, f(x), crosses the x axis

Answer 33

so we have 1 root already, that is x=0

Answer 34

okay

Answer 35

\[x(x^2-4x-12) = 0\] this is when either x = 0 or when \[x^2-4x-12 = 0\]

Answer 36

okay

Answer 37

since the order of this is 2 we need two values of x that make this true

Answer 38

So its a.

Answer 39

\[x^2-4x-12 = (x+a)(x+b)\]

Answer 40

-12 = a*b
-4x = a*x+b*x
x^2 = x*x

Answer 41

since x is in all terms of the 2nd equation you can divide the whole thing by 'x' and not cahnge it so it becomes:
-4 = a + b

Answer 42

so you have:
-12 = a*b
and
-4 = a+b

Answer 43

what will a and b equal?

Answer 44

-2 and 6 right?

Answer 45

-4 = -2 + 6?

Answer 46

now plug a and b into the equation above to get \[x^2 -4x -12 = (x-6)(x+2) = 0\]

Answer 47

now what values of 'x' makes each root = 0?

Answer 48

x = 0 or -2 or 6 to make f(x) = 0