Question

2.Determine how many, what type, and find the roots for f(x) = x^3 - 5x^2 – 25^x + 125.

Answer 1

it is a polynomial of degree 3 so there are 3 roots, although two may be complex roots

Answer 2

you are supposed to find them and you have a lot of fives there, so maybe five is a root

Answer 3

oh my there is some kind of typo there i think

Answer 4

I tried to like factor it and do all that but I got (x - 5)(x^2 - 25x + 125) and then I don't know how to go on from there

Answer 5

\[f(x)=x^2+5x^2-25x+125\]

Answer 6

if you got one root then what remains is a quadratic
you can always find the roots of a quadratic, use the formula

Answer 7

you factored incorrectly is the problem

Answer 8

\[(x-5)(x^2-25)\] would be better

Answer 9

how did you get that?

Answer 10

synthetic division is easiest here
or you can just think about what it has to be
\[x^3+5x^2-25x+125=(x-5)(ax^2+bx+c)\] and \(a, c \) are totally obvious

Answer 11

since you have to have \(x^3\) then \(a=1\) and since you have to end up with \(125\) you see \(c=-25\) making it
\[(x-5)(x^2+bx-25)\]
then it will turn out that \(b=0\) giving you \((x-5)(x^2-25)\) or
\[(x-5)(x-5)(x+5)\] after you factor the second part

Answer 12

I just did the synthetic division but im not sure what im supposed to get...

Answer 13

English please lol

Answer 14

you should get as the bottom line

1 0 -25 0

Answer 15

1 -5 -25 125
5 5 0 -125
_____________________
1 0 -25 0

Answer 16

that is what the synthetic division would look like if you are dividing by \(x-5\)
the bottom row gives you the quotient of \(x^2-25\)

Answer 17

oooohhh ok I get it. can you help with two more please?

Answer 18

kk

Answer 19

3.The following graph shows a seventh-degree polynomial:
graph of a polynomial that touches the x axis at negative 5, crosses the x axis at negative 1, crosses the y axis at negative 2, crosses the x axis at 4, and crosses the x axis at 7.
Part 1: List the polynomial’s zeroes with possible multiplicities.
Part 2: Write a possible factored form of the seventh degree function.

Answer 20

this Is only one but theres another question later, but I think its easy

Answer 21

the zeros you see from your eyeballs right?

Answer 22

the zeros are where the line touches the x-axis right?

Answer 23

touch or cross yes

Answer 24

how do I make a factored form?

Answer 25

list them

\[{-5,-1,4,7}\] so it will be\[(x+5)^a(x+1)^b(x-4)^c(x-7)^d\] we just need the exponents

Answer 26

it just touches at \(-5\) and bounced back up, so that exponent is even, no doubt it is 2

Answer 27

the exponents 2?

Answer 28

at \(-1\) it crosses
that exponent is odd, probably 1
now we are at
\[(x+5)^2(x+1)(x-4)^c(x-7)^d\]

Answer 29

the exponent on the factor of \(x+5\) is two yes
that is because it touches and bounces back up
doesn't cross
that is called the "multiplicity" of the zero

Answer 30

you see it does cross at 7 so that exponent is also 1
\[(x+5)^2(x+1)(x-4)^c(x-7)\]

Answer 31

oooh ok

Answer 32

and also you see that it crosses at 4, but it is flattened out somewhat there
doesn't cross like it does at -5 or at 7
that makes the exponent odd, but not 1, probably 3

Answer 33

that and the fact that you were told it is a poly of degree 7
we should surmise
\[(x+5)^2(x+1)(x-4)^3(x-7)\] which does have degree 7 as asked for

Answer 34

look at the top picture here to confirm
http://www.wolframalpha.com/input/?i=%28x%2B5%29^2%28x%2B1%29%28x-4%29^3%28x-7%29

Answer 35

ok thank you but one question...if the line touches at -5, -1, 4, and 7 how come they are the opposite sign? is it supposed to be like that?

Answer 36

if you have
\[x-3=0\] that meas
\[x=3\] right?

Answer 37

oh ok I feel stupid lol I get it now. one more question?

Answer 38

if you want to solve
\[(x-3)(x+4)=0\] that means
\[x=3\] or \[x=-4\]
conversely if you know the zeros are \(-4\) and \(3\) you know it factors like
\[(x-3)(x-4)\]

Answer 39

k

Answer 40

i made a typo on the last post
ignore it

Answer 41

4.Without plotting any points other than intercepts, draw a possible graph of the following polynomial:
f(x) = (x + 8)^3(x + 6)^2(x + 2)(x – 1)^3(x – 3)^4(x – 6).

I know its like the last one and I get how to graph the zeros but I don't know what the line is supposed to lo like.

Answer 42

@satellite73

Answer 43

crosses at -8
touches but does not cross at -6
crosses at -2
crosses at 1, flattened out a bit
touches but does not cross at 3, flattened out a lot
crosses at 6

Answer 44

take a look here
http://www.wolframalpha.com/input/?i=+%28x+%2B+8%29^3%28x+%2B+6%29^2%28x+%2B+2%29%28x+%E2%80%93+1%29^3%28x+%E2%80%93+3%29^4%28x+%E2%80%93+6%29.

Answer 45

thanks but im not allowed to use a thing that graphs for me. how would I know if I didn't use it?

Answer 46

by what i wrote above
lets take them one at a time

Answer 47

one factor is \( (x + 8)^3\)
the zero is \(-8\) and the multiplicity is 3
3 is odd, so it crosses the axis there
but because it is 3 and not one, it is flattened out a bit as it crosses

Answer 48

next factor is \((x + 6)^2\)
the zero is \(-6\) and the multiplicity is 2
because 2 is even, it touches but does not cross there

Answer 49

etc

Answer 50

ooohhhh ok thanks :)

Answer 51

@satellite73 I need help plz