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Part A
Basically isn't this just asking to factor the equation?
(x +2) * (x +3)

Part B
(x -2) * (x -2)

but how are you getting these...

I guess you don't know how to factor.

sadly no...

so in this case, a= 2 b=5 and c=6 ???

If you are referring to part A then a would equal 1 and you have the rest correct.

sorry, i missed that.. i understand that part. thank you! but can you explain part B?

Okay for part B the equation is
x2 - 4x + 4
and a =1 b=-4 and c=4

so its just like part a?

Yes they both require you to factor the equation

seems like i was stressing over nothing! somehow you made it clear to me, thank you so much

im having trouble finding the gcf

im new to this, its hard with no teacher or anyone to explain..

thats the part i was mostly confused on.

Well that's my answer. I don't know if it the correct answer but it's the best I could come up with.

thank you, do you mind if you can walk me through some other ones?

I don't see a common factor for part c either

Thanks UsukiDoll :-)

Thank you both.

you're welcome :)

u r welcome

can you help me on 3 more questions?

Three more??? Well okay. :-)

yes.. but this is the last time having to deal with this topic...

for part a.... a=1 b=2 c=-2
not 100% sure...

Part A
a=1 b=2 c=-1 d=-2

You are now factoring a cubic equation

totaly just misssed the "x" there.. sorry about that

totally**

factor by grouping for the first part
rational root test for the second part

im new that this.. @UsukiDoll

and it's been a while since I factored a cubic equation

sorry x.x

B should be
\[(x-2)^2 \]
because that is in square of a linear expression formio

(x-2)^2 and (x-2)(x-2) mean the same thing.

For (x-2)^2 that exponent number is 2 meaning write (x-2) 2 times
(x-2)(x-2) .

how did you get that

I was talking to Shalante earlier...

gotcha! i wish i could understand this, the first problem i got, but now this is totally different

I know... I got distracted... one sec while I cool off before going berserk.

okay, thank you for your help!

UsukiDoll it seems you know how to factor a cubic equation?

\[f(x) = x^3+2x^2-x-2\]

You mentioned grouping

I can't do anymore of this... I know it, but I'm being picked on.

no one is picking on you...

\[\large f(x) = x^3+2x^2-x-2 \]
through factor by grouping we have
\[\large x^2(x+2)-(x+2)\]

\[\large f(x) = (x+2)(x+1)(x-1) \]
that's the final factored form

thank you so much for explaining this to me.

so now we have to find our 0's... we just have to solve for x in 3 cases
x+2 = 0, x+1=0, x-1 =0

did you solve for x first in the previous part?
there should be 3

2,1,1????

close....
solve for x
x+1= 0
subtract 1 on both sides

some how im getting 2,1 and then -1 but im over thinking this and getting more confused

it's correct now
x =2,1,-1

-2 -1 and1

oh wait I missed. ugh yeah wolf is right -2 -1 1

I overlooked by accident.

you did the factoring though

made the same mistake.

It was an accident... geez.

but you factored it correctly Usuki :-)

its late, im sure we all make mistakes.. its okay.. your doing a good job!

well I'm gonna get going

I'm out too. the nitpicking towards me is too much.

nobody is nitpicking, your helping me out alot