Question

Factoring help

Answer 1

2x^2-36x-38=7

Answer 2

wait no typo

Answer 3

2x^2-36x-38=y

Answer 4

\[x=\frac{ 18+\sqrt{2(y+ 200)} }{ 2 } ,\frac{ 18-\sqrt{2(y+200)} }{ 2 }\]

Answer 5

uhh I wasn't taught how to solve it with that method..

Answer 6

lol i think ur using some kind a calculator

Answer 7

Nah

Answer 8

the question was to factor the function, right?

Answer 9

yeah find the roots algebraically

Answer 10

\[y=2x^2-36x-38\]

Answer 11

well is it
2x^2- 36 - 38 = 0 ?

Answer 12

yeah

Answer 13

* -36x

Answer 14

you can simplify it a bit by dividing through by 2 giving

x^2 - 18x - 19 = 0

it will factor - you need 2 numbers whose product is -19 and whose sum is -18

can you think of them?

Answer 15

\[y=2(x^2-18x - 19)\]

Answer 16

calculator says that

Answer 17

^^

Answer 18

the numbers are -19 and + 1
so we have

( x - 19)(x + 1) = 0

Answer 19

those are the 2 factors

Answer 20

So it would be (x-19)(x-1) ?

Answer 21

no x+1 not x-1

Answer 22

if you what the roots then you solve the 2 equations

x - 19 = 0 and

x + 1 = 0

Answer 23

* want

Answer 24

- but the factors are x - 19 and x+ 1

Answer 25

ohh I totally forgot about that part

Answer 26

kk so i got x=19 and x=-1

Answer 27

yae if 2 things are multiplied together and are equal to 0 then either one can be = 0.

Answer 28

right

Answer 29

so the roots are 19 and -1 ?

Answer 30

yes

Answer 31

Alrighty THANK YOU so for much for help !!!! @welshfella and @alivejeremy
I really appreciated it XD !

Answer 32

yw

Answer 33

np