Question

x^2+10x+16 x+8
______________ / ________
x^2-6x-16 x^2-64

Answer 1

so i need help factoring this equation

Answer 2

first things first rewrite it so that it's multiplied
division = multiplication by the reciprocal

Answer 3

so do the keep change flip thing?

Answer 4

for example dividing by 3/2 is the same t hing as multiplying by 2/3

Answer 5

I don't remember the keep change thing

Answer 6

wait were u talking about changing it to multiplication?

Answer 7

yeah.

\(\large \frac{x^2+10x+16 }{x^2-6x-16 }\div \frac{x+8}{x^2-64}=\frac{x^2+10x+16 }{x^2-6x-16 }*\frac{x^2-64}{x+8}\)

Answer 8

yeah thats what i was talking about

Answer 9

nowe we start factoring and canceling

Answer 10

yeah thats what i have a hard time with the factoring

Answer 11

same

Answer 12

factor \((x^2-64)\)

Answer 13

this might help you out \((a^2-b^2)=(a+b)(a-b)\)

Answer 14

that be like (x+8)(x-8) right?

Answer 15

yas

Answer 16

what about \(x^2+10x+16\)

Answer 17

(x+8)(x+2) ?

Answer 18

correct. so let's update our fraction
\(\huge\frac{(x+8)(x+2)}{x^2-6x-16 }*\frac{(x+8)(x-8)}{x+8}\)

Answer 19

ok so now we need x^2-6x-16 but im having trouble with that one

Answer 20

list out the factors of 16

Answer 21

1,2,4,8,16
what combination of these can we use to get -6?

Answer 22

2 and 4? but that doesn't multiply into 16

Answer 23

8 and 2
the way it's written the pairs match outwards
1-16
2-8
4-4

Answer 24

so it would be (x+8)(X-2)?

Answer 25

that would be positive 6, we want -6
the way I think about it is 2-8=-6
or -8+2=-6

Answer 26

ohh oops ok so (x-8)(x+2)

Answer 27

looks gewd

Answer 28

time to reupdate the fraction

Answer 29

\[\huge\frac{(x+8)(x+2)}{(x-8)(x+2) }*\frac{(x+8)(x-8)}{x+8}\]

Answer 30

so i can cross out (x+2) right?

Answer 31

yeah

Answer 32

which (x+8 do i keep?

Answer 33

whichever you want. conceptually, it's one long fraction

Answer 34

does it matter actually? that should be my answer right?

Answer 35

\(\huge\frac{(x+8)\cancel{(x+2)}}{(x-8)\cancel{(x+2)} }*\frac{(x+8)(x-8)}{x+8}\)
\(\huge\frac{(x+8)(x+8)(x-8)}{(x-8) (x+8)}{}\)

Answer 36

then we can cancel an x+8 and x-8

Answer 37

so yeah the answer would just be x+8

Answer 38

yep

Answer 39

can u help me with another one?

Answer 40

sure

Answer 41

\[\frac{ 6x^2-32x+10 }{ 3x^2-15x}\div \frac{ 3x^2+11x-4 }{ 2x^2-32 }\]

Answer 42

ew I never learned how to do the ones with a>1

Answer 43

let me google it

Answer 44

for starters we can factor \(2x^2-32\)

Answer 45

factor 2 out.
\(2(x^2-16)\)

Answer 46

2(x+8)(x-8)?

Answer 47

4*

Answer 48

16=4^2

Answer 49

+4 or -4?

Answer 50

both

Answer 51

a^2-b^2=(a+b)(a-b)
x^2-16=x^2-4^2

Answer 52

what? lol how would u write it?

Answer 53

\(a^2-b^2=(a+b)(a-b)\)
\(x^2-4^2=x+4(x-4)\)

Answer 54

wait but then where would the 2 go?

Answer 55

what we have is 2(x+4)(x-4)
we can either leave it like that or write it as \(((2*x)+(2*4))(x-4)=(2x+8)(x-4)\)

Answer 56

OHHHHHHHHHHH!!!!!!

Answer 57

woah, forgot to google the other thing, give me 2 minutes

Answer 58

okieieee

Answer 59

\(ax^2+bx+c=0\)
Step 1: Find two numbers that multiply to give ac (in other words a times c), and add to give b.
step 2: rewrite the middle

Answer 60

\(6x^2-32x+10\)
ac=6*10=60
b=-32

not sure how that works out here
let's do the thing we did before with \(3x^2-15x\)

Answer 61

ok that be 3(x-5) right?

Answer 62

I'd factor out 3x

Answer 63

how?

Answer 64

\(3x(\frac{3x^2}{3x}-\frac{15x}{3x})\)

Answer 65

so that comes out to 3x(x-5)

Answer 66

ohhhhh i for got the x

Answer 67

yus

Answer 68

uhh let's see where we're at

Answer 69

\(\huge \frac{ 6x^2-32x+10 }{ (3x)(x-5)}\times\frac{2(x+4)(x-4) }{ 3x^2+11x-4 }\)

I'm going to cheat for the other 2

Answer 70

okayy

Answer 71

\(6x^2-32x+10=2 (x-5) (3 x-1)\)

Answer 72

\( 3x^2+11x-4=(3 x-1) (x+4)\)

Answer 73

so x-5 crosses out and so does x+4

Answer 74

btw I'm using this http://www.wolframalpha.com/input/?i=factor+3x%5E2%2B11x-4
\(\huge \frac{ (x-5)(3x-1) }{ (3x)(x-5)}\times\frac{2(x+4)(x-4) }{ (3x-1)(x+4) }\)

Answer 75

oh an so does 3x-1

Answer 76

\(\huge \frac{ \cancel{(x-5)}\cancel{(3x-1)} }{ (3x)\cancel{(x-5)}}\times\frac{2\cancel{(x+4)}(x-4) }{ \cancel{(3x-1)}\cancel{(x+4)} }\)
looks like the canceling leaves us with
\(\huge \frac{2(x-4)}{3x}\)

Answer 77

that website helps factor?

Answer 78

it does everything. calculus, probability whatever
you say "factor XXXXXXXXXXXXXX" it factors
you say "derive YYYYYYYYYYYY" it derives

Answer 79

:o this is amazing

Answer 80

do u have to factor with subtraction and addition?

Answer 81

I know! I'm sure there are other calculators online that do this too
but if you want my honest opinion, you should check out a factoring guide on youtube

what do you m ean

Answer 82

like for this...
\[\frac{ 6y-4 }{ y^2-5}+\frac{ 3y+1 }{ y^2-5 }\]

Answer 83

it should work you just have to learn how to use parentheses. watch
```
simplify ((6y-4)/(y^2-5))+((3y+1)/(y^2-5))
```
it's messy
http://www.wolframalpha.com/input/?i=simplify+%28%286y-4%29%2F%28y%5E2-5%29%29%2B%28%283y%2B1%29%2F%28y%5E2-5%29%29

btw that's in common base so you can just rewrite it as \(\large \frac{6y-4+(3y+1)}{y^2-5}\)

Answer 84

is that the answer? do u combine like terms?

Answer 85

yeah, always

Answer 86

9y-3

Answer 87

i see wb this one?

Answer 88

dammm bby
getting a bit out of hand

Answer 89

jk not doing anything of value right now anyhow

Answer 90

\[\frac{ 3 }{ 8x^3y^3 }-\frac{ 1 }{ 4xy }\]

Answer 91

lol sorry after this one i have like 2 more questions so im like merp
wanna finish

Answer 92

factor the bottom left denominator

Answer 93

yea, we'll finish

Answer 94

how do i do that

Answer 95

there are 2 ways of thinking about it. we want the least common multiply of 4xy and 8x^3y^3

what my brain instinctively says is factor 4 x and y
\(8x^3y^3=4xy(2x^2y^2)\)

Answer 96

then we know that if we multiply the right hand side by \(\large \frac{2x^2y^2}{2x^2y^2}\) (which is effectively 1, we'll have a common denominator.

Answer 97

so the bottom right side would turn into 8x^3y^3 and the top would be 2X^2y^2?

Answer 98

yeah