Question

More factoring.

Answer 1

you need the answer?

Answer 2

Yes......

Answer 3

Ok let me solve it

Answer 4

I need to learn how to do this... Ik what gcf is, but I can't find any numbers to divide 11 with

Answer 5

I see my mistake now....its 11.

Answer 6

the gcf is 11

Answer 7

The answer is : 11yz^3 ( z^4-3x^4y^3)

Answer 8

how did you get that.?

Answer 9

you need to factor out the commen term

Answer 10

Yes.. but...

Answer 11

11yz^7-33x^4y^4z^3

Answer 12

I'm always confused on the stupid exponets

Answer 13

Yerp

Answer 14

you add all the exponents together after you factor out the commen terms

Answer 15

if that makes sense

Answer 16

yes but...its confusing to the point where there are so many different ones.

Answer 17

Im' confused on the last one..

Answer 18

still looking for help?

Answer 19

yes

Answer 20

bcs im looking at the last y.

Answer 21

you said y^3 but there are 4 ys.

Answer 22

SO did you put the other y at the beginning?

Answer 23

This is why math isn't my strongest subject...

Answer 24

Right, I'm going to follow the logic of the person above.

okay so we have \( \ \ 11yz^7-33x^4y^4z^3\)

you see we have got two terms \(11yz^7\) & \(33x^4y^4z^3 \) which include two real numbers in \(11\) and \(33\) which can both be factored \(=11yz^7-11\times 3x^4y^4z^3\)

therefore we get \(11\left(yz^7-3x^4y^4z^3\right)\)

okay so insidet he brackets for the \(y\) part we know that \( a^{b+c}=a^{b} \times a^{c} \) so \(y^4=y\times y^3\)

so we get \( y\left(z^7-3x^4y^3z^3\right) \rightarrow 11\times y\left(z^7-3x^4y^3z^3\right) \)

Can you try and work out the \(z\)'s ?

Answer 25

I understand all of that part. The z's would be 3 and 7, You took 3 from 7. which gave you 4

Answer 26

right! then what's wrong?

Answer 27

The y.

Answer 28

well both terms have got only ONE \(y\) in common

\( 11\left(\color{steelblue}{y^1}z^7-3x^4\color{steelblue}{y^4}z^3\right) \)

i explained above

Answer 29

well. Then ig it makes sense now.

Answer 30

Thanks for helping me.