Question

Factorizing By Quadratics

Answer 1

\[2x ^{2}-2x\]

Answer 2

What the greatest common factor? ALWAYS factor out any GCF first.

Answer 3

Annnnnnd... that will pretty much be it. :)

Answer 4

I got it lol. Wasn't thinking properly.
\[2x\left( x-1 \right)\]

Answer 5

Exactly... :)

Answer 6

Al right this one is \[-6x-1x ^{2}+9\]
I think I have to move 1x^2 over to the beggining and solve normal. So it will be
\[-1x ^{2}-6x=9\] and solve normally?

Answer 7

.

Answer 8

Where did that = sign come from? You mean a + there, right?

And then I would factor out the -1 first, so that you are dealing with a positive leading coefficient when you do the factoring (that isn't strictly necessary, but will make it easier to factor).

Answer 9

\[\left( x+3 \right)\left( x-3 \right)\]

Answer 10

Did you CHECK YOUR FACTORING WITH MULIPLICATION? hmmmm??

(Didn't I give you that speech, or do I need to give it again!? :P lol)

Answer 11

If i am right a postive times a negative = a negative... *whistles and looks around* What i did wrong :P

Answer 12

\(-1x ^{2}-6x+9=-(x^2+6x-9)=-(..?..)(..?..)\)

Notice you factor out the -1, but it doesn't "go away", it's still there.

Are you sure this one factors? Are any of them prime?

Answer 13

\(\Large \left( x+3 \right)\left( x-3 \right)=x^2-9\neq -x ^{2}-6x+9\)

Answer 14

Honestly I'm lost O.o so... (x-3) (x-3)

Answer 15

ok, well.... your proposed factoring above was wrong. I showed you that, by checking it with multiplication. :) you don't get back to the original expression.

so back to square one: like I said, start by factoring out that -1, that way you aren't trying to factor with a negative leading coefficient.

\(\Large -1x ^{2}-6x+9=-(x^2+6x-9)=-(..?..)(..?..)\)

But then I noticed, that does NOT factor. \((x^2+6x-9)\) is prime, because the only way to get the product of -9 in the 3rd term is with either 3*3 with opposite signs, or 1*9 with opposite signs... neither of which will sum to the +6 needed in the middle.

So this one does NOT factor, it's a "prime polynomial".

Answer 16

:(

Answer 17

No need for sad face... some polynomials are just prime. A lot are, in fact. :) just like integers, they don't ALL factor. :)

Answer 18

SO there is no answer to that question?

Answer 19

If, in the problem set, it is presented in such a way that ALL are supposed to factor, then that's an error... unless they JUST want you to factor out the -1, which is a possibility.

After that, there is nothing more to factor out; what you have inside the ( ) is a prime polynomial.

Answer 20

Maybe i wrote down the question wrong. Thanks very much for the help :)

Answer 21

It isn't that there is "no answer", it's that the polynomial is prime.

Most problem sets of this type that I've seen usually have instructions something like, "factor the polynomial completely; if the polynomial is prime, say so", leaving open the option that some are prime (because it's also important that you understand when that is the case! :)

Answer 22

Alright :)