Question

Factor 2x^2+6x+2

Answer 1

can not be factored in terms of integers

Answer 2

2(x^2+3x+1) so after you pulled out the two you start factoring using any method you learned, in this case when you try to factor our two numbers that multiply to equal 1 and add to equal three, there is no factor

Answer 3

you can now try the quadratic formula and see if that gives you the answer u want

Answer 4

how do I do that?

Answer 5

you can factor out of 2, but you can't factor it more once you do that.

Answer 6

what do you mean?

Answer 7

I mean once you pull the 2, \(\large\color{blue}{ 2x^2+6x+2 }\)
\(\large\color{blue}{ 2(x^2+3x+1) }\)
from this point, you can't factor more,
because \(\large\color{blue}{ x^2+3x+1 }\) is NOT factorable.

Answer 8

you can check this, by
\(\large\color{blue}{ x^2+3x+1 }\)

\(\large\color{blue}{ 3^2-4(1)(1)=9-4=5 }\)
the discriminant is NOT a perfect square.

Answer 9

so theres no answer?

Answer 10

there IS an answer, and it is \(\large\color{blue}{ 2(x^2+3x+1) }\) .

Answer 11

you factored as much as you could, pulled out a 2, but you can't factor it more.

makes sense?

Answer 12

hmm how do I find the zeros for that then?

Answer 13

you would use a different method. Completing the square or the quadratic formula.

Answer 14

you would use a different method. Completing the square or the quadratic formula.

Answer 15

I'm just going to post the whole problem, maybe it changes things?

Answer 16

I can't read it it is too small

Answer 17

actually let me see....

Answer 18

but if that is the problem how did you get 2x^2+6x+2=0 ?

Answer 19

from synthetic division

Answer 20

by what?

Answer 21

you mean like how? graphing the equation and finding a root

Answer 22

you divided the polynomial by one of it's roots?

Answer 23

I don't understand by dividing the function, `by what` have you obtained 2x^2+6x+2.

Answer 24

.5

Answer 25

i get what youre saying now

Answer 26

so now you have \(\large\color{black}{ 2(x^2+3x+1)=0 }\) right?

Answer 27

So you divided by (x-0.5), the root, when a zero is 0.5, and you got that.
\(\large\color{black}{ 2x^2+6x+2=0 }\)
If so, then it is right:)

Answer 28

that means that x=0.5 is one of the solutions, no lets go on solving for the other 2 solutions.
Tell me when you are okay to continue.

Answer 29

im okay

Answer 30

to continue

Answer 31

Sorry connection snapped.

Answer 32

\(\large\color{black}{ 2(x^2+3x+1)=0 }\)
\(\large\color{black}{ x^2+3x+1=0 }\)

So, the equation is \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{3}x+ \color{red}{1}=0 }\).

Fill in YOUR values. \(\bbox[8pt, lightyellow,border:8px solid black]{\LARGE \LARGE{x=~} \huge{ \frac{-\color{magenta}{3} \pm\sqrt{ \color{magenta}{(3)} ^2-4 \color{blue}{(1)} \color{red}{(1)}}}{2 \color{blue}{(1)}} } ~ }\)

Answer 33

can finish, or need help?

Answer 34

let me try, one sec

Answer 35

sure

Answer 36

i dont know what to do about the square root of 5

Answer 37

just leave the way it is, if you want the exact solution, because this is probably that you want.

Answer 38

\(\bbox[8pt, lightyellow,border:8px solid black]{\LARGE \LARGE{x=~} \huge{ \frac{-\color{magenta}{3} \pm\sqrt{ 5}}{2 \color{blue}{(1)}} } ~ }\)

Answer 39

\(\bbox[8pt, lightyellow,border:8px solid black]{\LARGE \LARGE{x=~} \huge{ \frac{-\color{magenta}{3} \pm\sqrt{ 5}}{2} } ~ }\)