Question

Can anyone teach me on factoring? Plssssssss.......

Answer 1

And rational expressions

Answer 2

You are better post the particular problem and practice on it. It will help you know how to master the method.

Answer 3

ok sure

Answer 4

\[12x^4y^5-18xy^6\]

Answer 5

Its easy when i do special products

Answer 6

But doing the other way round I just dont know how. I get screwed up.

Answer 7

I saw uncle Rhaukus is writing, let's wait for uncle, ok?

Answer 8

Fully expanding the terms makes it a bit clearer
\[12x^4y^5−18xy^6= 2\times2\times3\times x\times x\times x\times x\times y\times y\times y\times y\times y\\\qquad\qquad\qquad\qquad-2\times3\times3\times x\times y\times y\times y\times y\times y\times y\]

Answer 9

Woooow!!

Answer 10

to factor the expression , we have to pull out the common factors,
the common factors are 2, 3, x, y, y, y, y, y,

Answer 11

why so many y?

Answer 12

@UnkleRhaukus It is perfect way to do. However, it's so.... basic. Please, give him teenager step, not baby's one like that.

Answer 13

\[12x^4y^5−18xy^6= 2\times2\times3\times x\times x\times x\times x\times y\times y\times y\times y\times y\\\qquad\qquad\qquad\qquad-2\times3\times3\times x\times y\times y\times y\times y\times y\times y\\~\\\qquad\qquad=2\times3\times x\times y\times y\times y\times y\times y(2\times x\times x\times x-3\times y)\]

Answer 14

Wait what? Its confusing...

Answer 15

\[12x^4y^5−18xy^6 = (6\times x\times y^5\times2\times x^3 -6\times x\times y^5\times3\times y)\]

Answer 16

There's actually so many types of factoring, and I screw at all of them.

Answer 17

now use
\[ab+ac=a(b+c)\]
where \(a\) is the product of all the common facors

Answer 18

you mean 2 is the common factor?

Answer 19

2 is one of the common factors

Answer 20

Because 12 and 18 are divisible by 2.

Answer 21

yeah so even x and y are common factors?

Answer 22

you could factor the 2 out in a first step if you want to
\[12x^4y^5−18xy^6=2(6x^4y^5-9xy^6) \]

Answer 23

yeah both of the terms have factors of x and y, in fact both have at least 5 factors of y in them, so you can factor out xy^5

\[=2(6x^4y^5−9xy^6)\\=2xy^5(6x^3−9y)\]

Answer 24

after we have factored out 2xy^5, there is one more common factor in the terms in the parenthesis ,

Answer 25

6 and 9 are divisible by 3 right?

Answer 26

thats right

Answer 27

But how would you put it out?

Answer 28

when 2 is already outside?

Answer 29

\[=2xy^5(6x^3−9y)\\=2xy^5(3\times2x^3−3\times3y)\\=3\times2xy^5(2x^3−3y)\]

Answer 30

\[6xy^5(2x^3-3y)\]

Answer 31

am i right?

Answer 32

That's right \[\checkmark \]

Answer 33

can you give me a question and I'll try to answer it?

Answer 34

what if we had
\[4ab^2+40a^2\]
how can we factorise it ?

Answer 35

Wait. Let me try on my scratch first.

Answer 36

ill give you a clue, there are only two common factors

Answer 37

\[4a(ab^2+10a)\]

Answer 38

that is so close

Answer 39

wait

Answer 40

\[4a(b^2+10a)\]

Answer 41

YESSSS !!

Answer 42

Let me try to solve for this one... Its quite hard.

Answer 43

210a^14-180a^8c

Answer 44

oh no....Im screwed again.

Answer 45

look at the number terms first, if it helps

210 and 180

both have factors of 10 right?

Answer 46

i used 3 as factors

Answer 47

i guess i was wrong....

Answer 48

yeah 3 is a factor also

Answer 49

the largest common factor between 210 and 180 is 10x3 = 30

Answer 50

\[30a^8(7a^6-6c)\]

Answer 51

Not sure with my answer though.

Answer 52

THat is correct!!

Answer 53

Sorry if I bother you alot.....Im a freshmen in college taking up civil engineering....

Answer 54

really?

Answer 55

let me try x(m+n)-(m+n)

Answer 56

to check it, just go back the other way, and hopefully you'll get what you started with

\[30a^8(7a^6−6c)\\=30a^8\times7a^6+30a^8\times-6c\\
=210a^{8+6}-180a^8c\]

Answer 57

hint

x(m+n)-(m+n)

is the same as

x(m+n)-1(m+n)

Answer 58

(x-1)(m+n)?

Answer 59

Yes, Great stuff !

Answer 60

If you do not have enough time to teach me anymore its ok. I'll try to ask someone else.... Just tell me if you need to go.

Answer 61

Cause i might have alot of questions.

Answer 62

i can do one, more, but after that maybe you should close this post and start a new one

Answer 63

Oh? your going?

Answer 64

i can do one more

Answer 65

\[6a^2b^4-18a^2b^5-14a^4b^8\]

Answer 66

this one has three terms but the idea is the same

\[wx+wy+wz = w(x+y+z)\]

Answer 67

wait im still solving.....

Answer 68

\[2a^2b^4(3a-9b-7b^4)\]

Answer 69

Looks like Im wrong

Answer 70

almost, you did find the right common factor to factor out,

but

\[2a^2b^4(3a−9b−7b^4)=6a^3b^4-18a^2b^5-14a^2b^8 \]

which isn't quite the same as \(6a^2b^4−18a^2b^5−14a^4b^8\)

Answer 71

wait i gave the wrong question...

Answer 72

\[6a^3b^4-18a^2b^5-14a^4b^8\]

Answer 73

There it is....

Answer 74

the last term still isn't right

Answer 75

\[2a^2b^4(3a-9b-7a^2b^4)\]

Answer 76

yeah i know....

Answer 77

\[2a^2b^4(3a-9b-7a^2b^4)\\=2a^2b^4\times3a+2a^2b^4\times-9b+2a^2b^4\times-7a^2b^4)\\=(2\times3)a^{2+1}b^4+(2\times-9)a^2b^{4+1}+(2\times-7)a^{2+2}b^{4+4}\\
=6a^3b^4-18a^2b^5-14a^4b^8\]
look like you got it!

Answer 78

you sure?

Answer 79

i can't see any thing wrong with the working, (other than that extra parenthesis)

Answer 80

parentesis is part of it right?

Answer 81

If you should go tell me.... I'll ask help from solomonZelman

Answer 82

i just mean i typed it a bit wrong\[2a^2b^4(3a-9b-7a^2b^4)\\=2a^2b^4\times3a+2a^2b^4\times-9b+2a^2b^4\times-7a^2b^4\color{red})\\=(2\times3)a^{2+1}b^4+(2\times-9)a^2b^{4+1}+(2\times-7)a^{2+2}b^{4+4}\\
=6a^3b^4-18a^2b^5-14a^4b^8\]
this extra parenthesis

Answer 83

i'm pretty sure you got it right

Answer 84

Ok. Well maybe i should open a new question then....