Question

Really really need help with these factoring questions! Can anyone help explain?

Answer 1

anyone??

Answer 2

really? can you help explain how to do it?

Answer 3

i did the division but i dont know how to factor it to get one of those answer choices..

Answer 4

ok i usually do long division
which one you feel comfortable long or synthetic division

Answer 5

synthetic is easier for me. i got the answer x^2-2x-3

Answer 6

but now im not sure how to get one of those answer choices..

Answer 7

I attached a picture!

Answer 8

Can you see it up there?

Answer 9

ok after they find answer you have to factor that

Answer 10

Once you have x^2 - 2x - 3, just factor that as you would any quadratic.

Answer 11

I suck at factoring. haha thats what im having an issue with.

Answer 12

and i cant figure out how to factor it to get one of those answer choices.

Answer 13

you can either do back ward and
multiplying your multiple choices

Answer 14

with doing that, do i just plug in one of the answers for each x?

Answer 15

then solve?

Answer 16

ohhh i found another easy way
do you have calculator

Answer 17

yup

Answer 18

ok go to y and plug that your first equation which is with x cubic
and then go to table find all x value where is y 0

Answer 19

to factor x^2 - 2x - 3, you need to find two numbers that multiply to -3 (last term) AND add to -2 (middle coefficient)

those two numbers are -3 and +1 since -3*(1) = -3 and -3 + 1 = -2

Which means x^2 - 2x - 3 = (x-3)(x+1)

Answer 20

Do you want to go over quadratic factoring in general, or just get the answer and move on?

Answer 21

oh oaky, so then it would be C? Because they tell you x=5 so you just carry that?

Answer 22

well i will have other questions i can use as reference! that would be great if you could help explain factoring them.

Answer 23

Well, I can go over all the cases and give you the main ways of solving each, or we can just only do whats needed to solve your specific problems, up to you.

Answer 24

yeah might as well just do what i need to do, i dont have that much time. im trying to get this activity done.

Answer 25

this one is a little different..

Answer 26

I feel like i know how to do it i just cant remember.

Answer 27

do you just switch the value? like positive or negative?

Answer 28

in that case it would be D, but im not sure.

Answer 29

Well, a "factor" is going to be written using the opposite sign of what the "root" is. So if you have x = 6 to be a root, then (x-6) is a factor, and so forth

Answer 30

thank you!! okay again with this one, i dont know why i just cant remember what to do. i know its simple though.

Answer 31

you have to write like (x + 1)
so you have -6, -2 , 3
when you have to write as a bracket with x so you have to do opposite of that number like
(x + 6)
because they gave you zeroes ( x + 6) = 0
when you solve for x you should get your zero which is x = -6

Answer 32

for last one first you have to multiply
can you do that

Answer 33

You would just be foiling out each group at a time. Like, you can start with trying to foil out (x-1)(x+2) and see what you get with that. After you get your answer with that, thn multiply it by (2x+2). Just go one step at a time. Would you be able to multiply out (x-1)(x+2)?

Answer 34

ohhh you distribute. okay let me try and figure that out.

Answer 35

alright i got 2x+2x-x-2

Answer 36

wait, that first one should be x^2 shouldnt it be

Answer 37

Yes, x^2. Now just combine the two midle like terms.

Answer 38

so then that leaves you with (x-1)(x^2+x-2)?

Answer 39

You've taken care of the first two factors, (x-1)(x+2) and we found that to be x^2 + x - 2. We don't need those two anymore now. The last factor we have to multiply out is now (2x+2). So you need to try to do (x^2+x-2)(2x+2)

Answer 40

the answer i got was 2x^3+2x^2+2

Answer 41

thats not an option hahaha ugh.

Answer 42

because that is not right :(

Answer 43

Alright, so let's do this one each step at a time. So the first group has 3 terms and the 2nd has 2 terms. Multiplied out, I will end up with 6 terms that I need to combine. So these are the multiplications we would get:
x^2 * 2x = 2x^3
x^2 * 2 = 2x^2
x * 2x = 2x^2
x * 2 = 2x
-2 * 2x = -4x
-2 * 2 = -4

That's all 6 of the multiplications you needed to do. If we combine it all together we have:

2x^3 + 2x^2 + 2x^2 + 2x - 4x - 4
2x^3 + 4x^2 - 2x - 4

Answer 44

if you do like box method then you cannot miss any number to multiply