#Why is the assignment saying that this is wrong?
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can you show your work?
yeah
or no actually the assignment says to factor out the common factor, have you done that?
do you know what it means to factor out a common factor?
sort of, you are indeed finding the GCF of the terms, but you are then going to factor out the GCF
so for example
if i have 3x^2+6x, what would be the GCF of both the terms?
2
do you know what a GCF is?
greater common factor
yep
so were looking for the greatest number that can divide 3x^2 and 6x evenly
does that make sense?
ohhhh
wait but is there a gcf for this???
because we cant divide 3 with 2 or 3 with 2
sorry i am slow with math
its all good dw
were looking for the greatest number that can divide 3x^2 and 6x evenly.
to do that we can first notice that 3x^2 and 6x can be divided by x evenly
now if we take a look at the coefficients we can notice that 3 divides 3 and 6 evenly too
thus, the GCF would be 3x
does that make sense or does anything need elaboration upon?
all the numbers can be divided by x even if 2 dosent have x?
where are you getting 2 from?
but there is no common gcf here except 1
well were looking at (x+3)^2-4(x+3)
the GCF can also be a liner factor too
lets take a look at the two terms
(x+3)^2 and -4(x+3)
do you see anything that can divide both (x+3)^2 and -4(x+3) evenly?
hint: ||it is going to be a liner factor||
(x+3)
alr 
wait so this is still correct but I was not doing what the question was asking
what would this be called then?
this form?
what your doing there is putting it into standard form
AHHH okay
alright so now that we know the GCF, all we need to do is to factor it out
okie let me do it and I will see if I get a right answer bwahha
๐
Btw you are so good at explaining
thx :)
thank you so much for taking time to help
no prob, its what im here to do
I am trying to figure out how to factor (x+3) bwahaha
ok, so when factoring were going to put parenthesis around the expression we want to factor
in this case we can do this:
( (x+3)^2-4(x+3) )
now were going to multiply the outside of the parenthesis by our GCF
that will get us
(x+3) * ( (x+3)^2-4(x+3) )
at the same time were going to divide the terms inside of the parenthesis by the GCF
that will yield
(x+3) * ( (x+3)^2/(x+3) -4(x+3)/(x+3) )
now simplify the inside, what will that get you?
(also if online notation is a bit overwhelming i can use the texit bot if you want)
ok, so when factoring were going to put parenthesis around the expression we want to factor
in this case we can do this:
$$\left(\ \ \left(x+3\right)^{2}-4\left(x+3\right)\ \ \right)$$
now were going to multiply the outside of the parenthesis by our GCF
that will get us
$$\left(x+3\right)\cdot\left(\ \ \left(x+3\right)^{2}-4\left(x+3\right)\ \ \right)$$
at the same time were going to divide the terms inside of the parenthesis by the GCF
that will yield
$$\left(x+3\right)\cdot\left(\ \ \frac{\left(x+3\right)^{2}}{\left(x+3\right)}-\frac{4\left(x+3\right)}{\left(x+3\right)}\ \ \right)$$
now simplify the inside, what will that get you?
Judgemental Snail
just try to simplify this part for now
(x+3)((x+3)-4)
I GOT IT
It cancels out the outher x+3
but becuase it square there is a additional one left
THat is okay, thank you soooooo much for helping me understand
I feel like this conversation we had def help me a lot
im glad :D
before you go though
i want to explain factoring using a literal definition instead of the process
you know the distributive property?
Yes I do
you can think of factoring as the opposite of that property
so for example, if we have
a(x+y),
we can use distributive property to get
ax+ay
like wise, we can factor out an a to get back what we just had before using the distributive property
does that make sense?
pretty much, factoring is the inverse of the distributive property ๐
just like how subtraction is the inverse of addition, ect..
Yes that makes much sense
Also quick question
sure
Do I ask a mod to change my name
because it's in chinese but I want to put it in english so people know what my name is
mhh im pretty sure you can edit your server profile
not sure if it requires nitro though
lemme check
they dont allow permission to change name in this server
alr looks good ๐
ok then i would DM modmail then, but not sure if it is permissible though -.-
if you do get redirected by the mods dw
if you go to the server list there should be a bot called "modmail" at the very top
should look like this
.solved
Post marked as solved by @blazing stirrup.
Use .unsolved if this was a mistake.