Question

i need help with gcf factoring

Answer 1

can someone like explain to me how to do this stuff?

Answer 2

\[4x^3y+8x^2y^2+2xy3\]

Answer 3

i had a phone confrence thingy with my teacher and she asked me to factor this

Answer 4

i failed horribly and i told her that i was doing something rn so we rescheduled it

Answer 5

GCF factoring is to factor out the largest factor possible.

Answer 6

Do you know how to find the greatest common factor between the numbers 4, 8, and 2?

Answer 7

2

Answer 8

what

Answer 9

oh

Answer 10

You are correct.

Answer 11

is that it? is that the answer!

Answer 12

So now we look at the variables.

Answer 13

it is 2 ....sorry

Answer 14

First, look at the numbers.
What is the greatest common factor of 4, 8, and 2?
That means: what is the largest number you can divide 8, 4, and 2 by evenly (with no remainder)? The answer is 2.

Answer 15

yup

Answer 16

so...2xy

Answer 17

? wut

Answer 18

So if we look at the x variable, if it is in all three terms, we look for the lowest exponent.

Answer 19

aint 2xy common in all the terms?

Answer 20

oh yeah

Answer 21

i get it

Answer 22

ssoo..answer is 2xy

Answer 23

the full answer?

Answer 24

2xy is what you factor out of each term.

Answer 25

\(\large 4x^3y+8x^2y^2+2xy3 =\)

\(= \large \color{red}{2} \times 2x^3y+\color{red}{2} \times 4x^2y^2+\color{red}{2}xy3\)

The GCF for numbers is 2, shown in red above.
Now let's look at variables.

Answer 26

yep its the complete answer.

Answer 27

You will have something like
2xy (__+__+__)

Answer 28

so thats it? its 2xy and we do nothing else?

Answer 29

...but its 2xy

Answer 30

No!

Answer 31

When you factor, you will have the gcf times a polynomial in parenthesis.

Answer 32

\(\large = 4xxxy+8xxyy+2xyyy \)

Below you see which variables are in common for all terms:

\(\large = 4xx\color{green}{x}\color{red}{y}+8x\color{green}{x}\color{red}{y}y+2\color{green}{x}\color{red}{y}yy \)

All three terms have xy in common.
That means the common factor is 2xy.
Now you need to factor it out.

Answer 33

oh i c

Answer 34

If you take out the common factor as mathstudent55 has shown you, what are you left with?

Answer 35

\(\large = \color{orange}2 \times 2xx\color{green}{x}\color{red}{y}+\color{orange}{2} \times 4x\color{green}{x}\color{red}{y}y+\color{orange}{2}\color{green}{x}\color{red}{y}yy\)

Answer 36

\[(2xy \times 2 =4xy) and(2xy \times 2=4xy)and(2xy \times 4=8xy)?\]

Answer 37

i mean 2xy times 1 is 2xy for the scnd on

Answer 38

idk what i just did

Answer 39

Now we keep the colors and move all the common factors to the beginning of each term:

\(\large = \color{orange}2\color{green}{x}\color{red}{y} \times 2xx+\color{orange}{2}\color{green}{x}\color{red}{y} \times 4xy+\color{orange}{2}\color{green}{x}\color{red}{y} \times yy\)

Now take out 2xy from each term, and put it up front.

\(\large = \color{orange}{2}\color{green}{x}\color{red}{y}(2xx + 4xy + yy)\)

Now we use exponents for the repeated variables.

\(\large = \color{orange}{2}\color{green}{x}\color{red}{y}(2x^2 + 4xy + y^2)\)

Answer 40

Can someone come and help me when they are done here, maybe? Please and thank you!

Answer 41

so thats our final answer?

Answer 42

TYSM I GET IT NOW

Answer 43

stacey can u medal math student?

Answer 44

Do you understand the following?
\[4x^{3}y=2xy(2x ^{2})\]
\[8x ^{2}y ^{2}=2xy(4xy)\]
\[2xy^{3}=2xy(y ^{2})\]

Answer 45

yup

Answer 46

can u give me an example so i can figure it out?

Answer 47

I just medaled mathstudent.

Answer 48

ty

Answer 49

brb getting a granola bar

Answer 50

Try \[6x ^{4}-12x ^{3} y+9x ^{2}y ^{2}\]

Answer 51

back @Stacey

Answer 52

\[(3x)\div6xxxx=2^4)and(3x)\div12xxxy=4x^3y)and(3x)\div9xxyy=3x^2y^2)so=3x(2^4-4x^3y+3x^2y^2)\]

Answer 53

\[so=3x(2^4-4x^3y+3x^2y^2)\]

Answer 54

@Stacey

Answer 55

????????

Answer 56

@mathstudent55

Answer 57

amirite?

Answer 58

\(\large 6x ^{4}-12x ^{3} y+9x ^{2}y ^{2}=\)

Rewrite every term fully factored, both numbers and variables.

\(\large = 2 \times 3 \times xxxx- 2 \times 2 \times 3 \times xxxy+ 3 \times 3 \times xxyy\)

Now what numbers and variables do all terms have in common?

Answer 59

Every term has a 3.
Every term has two x's.
The common factor is \(3xx = 3x^2\)

Now we group together 3x^2 in each term.

\(\large = 2 \times (3xx)xx- 2 \times 2 \times (3xx)xy+ 3 \times (3 xx)yy\)

Move the 3x^2 to the beginning of each term, and rename the repeated variables with exponents.

\(\large = (3xx) \times 2x^2 - (3x^2)4xy + (3 x^2) \times 3y^2\)

Answer 60

Now factor out the common factor, and keep the rest in parentheses.

\(\large = 3x^2(2x^2 - 4xy + 3y^2)\)

Answer 61

ok, gtg

Answer 62

@Stacey Great explanations above & thanks for the medal.

Answer 63

Sorry, I left for a bit.
mathstudent55 factored it correctly.

Answer 64

@fantageplayer You saw that x was in each term, so the next step is to look at the exponents
x squared was the lowest exponent, so \[x ^{2}\] is part of the gcf.

Answer 65

i dont get how to find the gcf of the exponents tho

Answer 66

@Stacey

Answer 67

like, how can the gcf be 3x^2 if 2 dosent fit into 3 evenly

Answer 68

Okay, I am going to do a binomial:\[24x ^{5}y ^{3}z ^{8}w + 18x ^{2}z ^{3}w ^{2}\]

Answer 69

6 is the coefficient.
We look at x: x^2 is the lowest exponent. It does not matter that 2 does not go into 5 evenly because we subtract exponents. We don't divide them.

Answer 70

y is not common to both terms.
We also have z^3 and w.

Answer 71

To find the parenthesis part divide by the gcf \[\frac{ 24x ^{5}y ^{3}z ^{8}w }{ 6x ^{2}z ^{3}w }+\frac{ 18x ^{2}z ^{3}w ^{2} }{ 6x ^{2}z ^{3}w }\]

Answer 72

\[4x ^{5-2}y ^{3} z ^{8-3}w ^{1-1}+3x ^{2-2} z ^{3-3}w ^{2-1}=4x ^{3}y ^{3} z ^{5}+3w\]

Answer 73

Factored we have: \[6x ^{2} z ^{3}w(4x ^{3}y ^{3} z ^{5}+3w)\]

Answer 74

i c what u did

Answer 75

give me one like that

Answer 76

so i can solve it

Answer 77

ill ask another question

Answer 78

\[12x ^{3}yz ^{2}w ^{5}-20x ^{3} y ^{3}w ^{2}\]