Question

-a^3 - 100a
would this be considered prime? meaning I could not factor it

Answer 1

beat me to it

Answer 2

\[-a^3 - 100a = -a(a^2 + 100)\]THis is far as you can actually factor it.

Answer 3

oh ok.. thank you

Answer 4

so I wouldnt have a common factor?

Answer 5

or would it be a

Answer 6

Well, you can factor out a GCF which would be -a.

Answer 7

so when worked out it looks like this:
-a^3 - 100a
-a(a^2+100) with a GCF of -a

Answer 8

Yes.

Answer 9

\[-(a)(a+10i)(a-10i)\]The *complete* factorization.

Answer 10

Yes. True. I just didn't do that because I'm not sure if you learned i yet.

Answer 11

this would be considered a perfect square

Answer 12

wouldnt the (a-10) be the same on both sides due to the -100a

Answer 13

What? Could you demonstrate what you mean?

Answer 14

(-a)(a+10)(a+10) because the positive and the negative will equal the -100a

Answer 15

nevermind they would both be (-)

Answer 16

Let's try to multiply all of this together then and see what we get?
\[(-a)(a^2 + 20a + 100) \rightarrow -a^3 - 20a^2 - 100a\]

Answer 17

See?

Answer 18

I mean (+)

Answer 19

What I wrote was correct. Don't worry about it.

Answer 20

-a^3-100a
-a(a^2 +100)
(-a)(a+10)(a+10)
when multiplied back
(-a) (a^2 + 100)
a^2 - 100a Right?

Answer 21

What? Not quite...

Answer 22

a^3 - 100a

Answer 23

\[(a + 10)(a + 10) = a(a + 10) + 10(a + 10) = a^2 + 20a + 100\]Not \(a^2 + 100\)

Answer 24

where did the 20a come from?

Answer 25

You have to FOIL it.
\[a \times a = a^2\]\[a \times 10 = 10a\]\[10 \times a = 10a\]\[10 \times 10 = 100\]

Answer 26

so how do I come up with a^3

Answer 27

when multiplied by the 10a?

Answer 28

sorry I am completely lost.. I thought I had it, but apparently I dont :(

Answer 29

You end up with something that isn't the correct answer which is what you were saying is correct.

Answer 30

Look. You sai that -a(a + 10)(a + 10) = -a^3 - 100a right?

Answer 31

yes

Answer 32

If you look at what I said above, (a + 10)(a + 10) = a^2 + 20a + 100

Answer 33

ok but I thought you had to multiple the (a +10)(a +10)

Answer 34

The:
\[-a(a^2 + 20a + 100) = -a^2 - 20a^2 - 100a\]See?

Answer 35

multiply

Answer 36

Are you confused about the (a + 10)(a + 10) part?

Answer 37

yup

Answer 38

Ok. Look. When you multiply two binomials together, you do something called FOIL.
First
Outside
Inside
Last
Multiply the terms in that order.
First terms are a and a.
(a)(a) = a^2
Outside Terms:
(a)(10) = 10a
Inside Terms:
(10)(a) = 10a
Last Terms:
(10)(10) = 100
Then you add them together.
a^2 + 10a + 10a + 100 = a^2 + 20a + 100

Answer 39

Do you get it?

Answer 40

ok so when I multiply back to check the answer I am not getting the original binomial. Am I not supposed to?

Answer 41

You're supposed to get the original answer back. THat's why I said that after \[-a(a^2 + 100)\]is unfactorable because you don't get that back.

Answer 42

So \[-a(a^2 + 100)\]is as far as you can get.

Answer 43

ok. so I have confused myself once again. making this harder than it really needs to be sorry.

Answer 44

Don't worry :) You're fine

Answer 45

this binomial would not be considered square correct?

Answer 46

Yes. A perfect square trinomial is in the form:
\[(a + b)^2 = a^2 + 2ab + b^2\]\[(a - b)^2 = a^2 - 2ab + b^2\]

Answer 47

ok thank you. was just checking

Answer 48

np :)

Answer 49

-a^3 – 100a The original expression to be factored. We notice that the
leading coefficient is (–a) and the constant term is an even number.
-a(a^2 – 100) The binomial cannot be factored any further and is considered complete.

Would this explanation be correct?

Answer 50

am I missing a step in the middle?

Answer 51

Well, I would say that the justification for the first step is wrong. The original expression has a Greatest Common Factor that can be factored out. Noticing how -a is in bother terms, I factored it out.

Answer 52

ok. the GCF would be the -a?

Answer 53

or would it actually be 10