Question

I will fan and medal.
f(x) = x3 + 5x2 - x - 5

Part A: What are the factors of f(x)? Please help :(

Answer 1

You must factor by grouping do you know how?

Answer 2

Yes. I finished the equation while waiting, but I don't think I did it quite right. I got to \[x ^{2}(x+5)-1(x-4)\] but I don't think that's right

Answer 3

Close but why -1(x-4)?

Answer 4

Because -5 and -x only have -1 in common. Do they not HAVE to have a common factor to solve?

Answer 5

If a multiply -1(x-4) I get -x+4 right?

Answer 6

Yeah. So that part of it is wrong? Ok so then it would just stay the same? x^2(x+5)(-x-5)?

Answer 7

No you were on the right track you still need to factor the( -x-5)

Answer 8

-1(what) would equal (-x-5)

Answer 9

1.2 Factoring: -x^3-5x^2+x+5

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: x+5
Group 2: -x^3-5x^2

Pull out from each group separately :

Group 1: (x+5) • (1)
Group 2: (x+5) • (-x2)
-------------------
Add up the two groups :
(x+5) • (1-x2)
Which is the desired factorization

(1-x^2)
Check : 1 is the square of 1
Check : x2 is the square of x1

Factorization is : (1 + x) • (1 - x)

Therefore (1 + x) • (1 - x) • (x+5)=0
solve for x you get x=-1,x=1,x=-5

Answer 10

So the factors for f(x) are (1+x), (1-x), and (x+5)?

Answer 11

\[(x+5)(x ^{2}-1)\]

Answer 12

How did you get that?

Answer 13

Also, how would I find the zeroes of this expression? Wouldn't it just be setting the factors equal to 0?

Answer 14

Above where you put -1(x-4) I said if you multply that out it is -x+4. Which is not -x-5.
But if you go with -1(x+5) you would get -x-5

Answer 15

So you hvae\[x ^{2}(x+5)-1(x+5)\]

Answer 16

Do you see this?

Answer 17

Yes. Alright, so now what do we do with x^2 and -1?

Answer 18

Distributive property

Answer 19

so (x^3+5x^2)(-x-5)?

Answer 20

No (x^2-1)(x+5)

Answer 21

Oh, alright. Then to get zeroes you set the factors to 0?

Answer 22

FActors to equal 0..Correct. Remember when doing x^2 it will have a positive and negative answer,

Answer 23

Wow, you've been such an amazing help! I have one last question. When we used the distributive property how come it didnt look like this