Question

A function is shown below:

f(x) = x3 + 5x2 - x - 5

Part A: What are the factors of f(x)? Show your work.

Part B: What are the zeros of f(x)? Show your work.

Answer 1

Factors are 1, 5, -1, -5, also multiplier of terms. Remember sign!

Answer 2

x3 + 5x2 - x -5 = 0
x2 ( x+ 5) - (x+ 5) = 0
Can you continue here?

Answer 3

Do you know how to factor this out?

Answer 4

Not really /:

Answer 5

wait so x3 + 5x2 - x -5 = 0
x2 ( x+ 5) - (x+ 5) = 0 is part b right @gebooors

Answer 6

Just because I need to learn better LaTeX I'll do it >.>

1. Split this 4 term polynomial into 2 groups

\(\ x^3 + 5x^2 - x -5 \rightarrow (x^3 + 5x^2)\) and \(\ (-x - 5) \)

Factor \(\ x^3 + 5x^2 \)

Notice they both don't have coefficient gcf, however, they do have variable gcf, "x"!

So if you factor a \(\ x^2\) from \(\ x^3 + 5x^2 \) you get \(\ x^2(x + 5) \)

Now can you factor \(\ (-x - 5)\) ?

Answer 7

so

it wold be -x(-5) ?

Answer 8

your response is in regards to part c right

Answer 9

Um no, there is no variable gcf or coefficient. So simply multiply it by -1. The reason why you're multiplying by -1 is because you want both terms to have a gcf. So if you use -1

You would get: \(\ -1(x+5) \) so \(\ -1(x+5) = -x - 5 \)

Now both terms have a gcf, (x+5) !

\(\ x^2(x+5) -1(x+5) \) now this becomes \(\ (x+5)(x^2-1) \)

Answer 10

You have factored it out, now to find the zero's, solve for x :)

Answer 11

What's part C, no were doing it together, then \(\ \color{blue}{you're} \) going to answer them with what you have solved. Should be easy, I'll be guiding you

Answer 12

if u say so

Answer 13

solve for "x"

Answer 14

lol, good one

Answer 15

Can you try to solve?

Answer 16

honestly, no, not even if a gun was held against me

Answer 17

o.o

K, well...

x + 5 = 0

and

\(\ x^2 - 1 = 0 \) now?

Answer 18

x2-1=0
+1 +1
x2=1

Answer 19

It's still x^2 = 1 continue, there's 1 more step (the inverse of the square is the square root)

Answer 20

k..

Answer 21

well u lost me, suprise suprise

Answer 22

wow, all of that to have the exact same answer except one less exponent.. YAY MATH

Answer 23

It won't always be like that...

Answer 24

x + 5 = 0

Answer 25

x=5

Answer 26

Oh wait a minute! you could have easily applied the differences of squares!

\(\ (x^2 - 1) = (x + 1)(x-1) \)

So you have (x + 5)(x-1)(x+1)

I'll just solve the zero's.

x + 5 = 0
-5 -5

x = -5

x - 1 = 0
+ 1 +1

x = 1

x + 1 = 0
-1 -1

x = -1


Overall: (-5,0) , (-1,0) and (1,0)

Answer 27

ok so thats part b right

Answer 28

Those are your x intercepts (zero's) yes.

Answer 29

http://prntscr.com/36txhv

Answer 30

Well yeah, we just used one of the factors and solved for the zero's. Maybe @Zale101 can clarify any question. I have to get off. Night!

Answer 31

thank you so much !!

Answer 32

Yw

Answer 33

Part C: What are the steps you would follow to graph f(x)? Describe the end behavior of the graph of f(x).

Answer 34

@tHe_FiZiCx99 there was one mre part /:

Answer 35

Sorry, I had to get off :/