Question

I have to get through everything involving factoring polynomials in one day! Can anyone help?

Answer 1

@phi @iGreen

Answer 2

@BVB4life @ganeshie8

Answer 3

Anyone? :/

Answer 4

Factor theorems??

Answer 5

Factor polynomials

Answer 6

For e.g like this factorise 2x^3-3x^2-11x+6

Answer 7

Right, polynomials.

Answer 8

Do u have a specific question?

Answer 9

Something like this.
-6x^4y-9x^3y^2+3x^2y^3

Answer 10

This thing has two variables! I don't think I can do this one

Answer 11

Do you know anyone who can?

Answer 12

No

Answer 13

phi! The savior! The light!

Answer 14

too much?..

Answer 15

There are a few ideas.

1) you need to know what a term is (things multiplied together)
example x^2y + x y^2 has two terms. the x*x*y and the x*y*y

2) if terms contain the same multiplier, you can factor them out. (opposite of distribute)
example: x*x*y + x*y*y
in the two terms both have an x*y. "factor it out" x*y ( x + y)

Answer 16

To tackle

\[ -6x^4y-9x^3y^2+3x^2y^3 \]

look at the numbers. (it helps to factor the numbers in to prime factors)
-6 is -2*3
-9 is - 3*3
3 is 3

is there the *same* number in all the terms ?

Answer 17

Yeah the 3

Answer 18

So our coefficient in our GCF is 3?

Answer 19

that means we can "factor out a 3"
i.e.
\[ -6x^4y-9x^3y^2+3x^2y^3 \\ 3(-2x^4 y-3x^3y^2+x^2y^3) \]
notice if you "distribute the 3" (multiply 3 times each term in the parens, we get back what we started with.

now let's do the letters

each term has at least 1 x. In fact they have multiple x's. Any idea how many x's you can "take out or factor from" each term ?

Answer 20

Well
x*x*x*x
x*x*x
x*x

Answer 21

all of them have at least x^2... right?

Answer 22

yes. so factor out x^2
Once you get used to the exponents (fast way of writing x*x), you will notice that you subtract 2 from each exponent to find what is "left behind)

what do we get if we factor out x^2 ?

Answer 23

Subtract right?

Answer 24

the exponents. and we put x^2 outside the parens, next to the 3

Answer 25

-2x^4y-3x^3y^2+x^2y^3
3x^2(-2x^2y-3xy^2+y^3)?

Answer 26

yes. now look at "y"

Answer 27

y
y*y
y*y*y

Answer 28

so just y?

Answer 29

yes

Answer 30

-2x^2y-3xy^2+y^3
3x^2y(-2x^2-3xy+y^2)?

Answer 31

yes. and that is as much as we can do.

notice the "letters" are easier than the numbers. For numbers you have to find the GCF. For the letters, you just count how many of the same letter is in each term.

Answer 32

Oh okay, but woah we got it wrong o.o

Answer 33

make sure the original is correct (no typos)

Answer 34

it says the answer is -3x^2y(2x^2+3xy-y^2)

Answer 35

So somewhere along the way we switched our signs or something..

Answer 36

oh, that is the same thing.

Answer 37

so what if we had a smaller one like 3a^2+30a?

Answer 38

there is no rule about the minus sign. But from
\[ 3x^2y(-2x^2-3xy+y^2) \]
we can factor out a -1

(that puts a -1 out front, and changes the sign of each term)

Answer 39

ohh I see what you mean

Answer 40

and to be clear,
\[ 3x^2y(-2x^2-3xy+y^2) \]
and
\[ -3x^2y(2x^2+3xy-y^2) \]
are equal valid.

Answer 41

Got you

Answer 42

notice if we distribute, both cases give back the original

Answer 43

3a^2+30a

Give it a try. First the numbers, then the letters

Answer 44

3*1
3*10?

Answer 45

yes

Answer 46

3(a^2+10a)?

Answer 47

yes. now do the "a"

Answer 48

3a(a+10)?

Answer 49

yes. as a check, distribute the "3a" (multiply each term inside the parens by 3a)
we get 3a^2 +30a
which matches.

Answer 50

okay! So the GCF is 3a?

Answer 51

yes

Answer 52

alrightttt!!!! Can I tag you a little bit later in factoring by another operation?

Answer 53

ok

Answer 54

Right on! Thanks!!