Question

factor completely x^2-9x+8

Answer 1

we want two numbers that multiply to get 8 but add to get -9, what are these numbers?

Answer 2

not sure

Answer 3

-8 -1

Answer 4

yes

Answer 5

does it matter which i put in which

Answer 6

ok now expand the middle term
x^2-9x+8 = x^2-x-8x+8
factor out what we can from the first two terms
x(x-1)-8x+8
now what can we factor from the 2nd two terms to make it x-1?
factor out -8
x(x-1)-8(x-1)
factor out x-1
(x-1)(x-8)=x^2-9x+8

Answer 7

I already gpt the answer

Answer 8

once you had the two numbers, because we have no coefficient on the x^2 term
we could have just said(x-1)(x-8)

Answer 9

but if you follow what I did, you can do any of these...

Answer 10

what about with a more complicated one, 6x^2 - 7xy -3x^2

Answer 11

x^2-8x-x+8
x(x-8)-1(x-8)
(x-1)(x-8)
x=1,x=8

Answer 12

what is 6x^2-3x^2?

Answer 13

3x^2

Answer 14

so you have
3x^2-7xy
do either of the two terms share anything?

Answer 15

y?

Answer 16

3x^2-7xy
x(3x-7y)

Answer 17

are there any y's in the first term 3x^2?

Answer 18

@farhan2h we don't need two people doing the same thing, I am trying to lead her to the answer...not just paste it.

Answer 19

i meant x sorry

Answer 20

ok so factor the x out and what do you have?

Answer 21

what the other guy said

Answer 22

yep

Answer 23

okay I get it

Answer 24

ok but notice that this is way easier than the other question, and if this had some other terms, it could have gotten very very ugly very very fast.

Answer 25

i have one with other terms

Answer 26

2u^2 + uv - 21v^2

Answer 27

ok so this is more like the first one,
we want two numbers that multiply to -42 and add to 1(the middle term has a 1 in front of it)
two such numbers are -6,7
so expand the middle term
2u^2+uv-21v^2 = 2u^2-6uv+7uv-21v^2
what can we factor out of the first two terms?

Answer 28

u

Answer 29

2u

Answer 30

okok

Answer 31

then we have
2u(u-3v)+7uv-21v^2
what can we factor out of the last two terms?

Answer 32

-21 v

Answer 33

not 21 but 7

Answer 34

you see why?

Answer 35

no

Answer 36

2u(u-3v)+7uv-21v^2
factor out 7u because each of the last two terms can be divided by 7u

Answer 37

if you had (21x+7) you can factor out a 7 not a 21
7(3x+1)

Answer 38

oh okay

Answer 39

I mean 7v

Answer 40

sorry

Answer 41

and we want it so that what we have left after we factor the last two terms looks like the thing in the first two terms

2u(u-3v)+7uv-21v^2
factor out 7v because each of the last two terms can be divided by 7v
2u(u-3v)+7v(u-3v)
let me know when you understand up to here.

Answer 42

i got it

Answer 43

ok notice now we have u-3v in both terms?

Answer 44

factor that out
2u(u-3v)+7v(u-3v) = (u-3v)(2u+7v)

Answer 45

this is the same thing as
2ux+7vx = x(2u+7v) but instead of x we have u-3v
so
2u(u-3v) + 7v(u-3v) = (u-3v)(2u+7v)

Answer 46

what we did

1) find two numbers that multiple to our coefficient on the last term, but add to the coefficient on the middle term

2) expand that middle term with out new numbers

3) factor what we can out of the first 2 terms

4) factor what we can out of the last 2 terms such that we are left with the same form as one of the factors from part 3)

5) then factor that thing out

I know this is not the best mathematical terms, but if you can do this with all the quadratics you get(that are factorable), and if you understand this your good...