Part A: Write the expression x2 + 5x + 6 as a product of two linear expressions. Show your work and justify each step. (5 points)
Part B: Rewrite x2 - 4x + 4 as a square of a linear expression. (3 points)
Part C: Do the expressions in parts A and B have a common factor? Justify your answer. (2 points)

- anonymous

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- wolf1728

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

- anonymous

yes but i have been having big problems trying to solve this things, i shouldn't be up crying over this....

- wolf1728

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

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## More answers

- anonymous

but how are you getting these...

- wolf1728

I guess you don't know how to factor.

- anonymous

sadly no...

- wolf1728

Okay let's take x2 - 4x + 4
in this case a=1 b=-4 and c=4
(where a, b and c are the factors of each term of the equation)
Do you understand that?

- anonymous

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

- wolf1728

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

- anonymous

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

- wolf1728

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

- anonymous

so its just like part a?

- wolf1728

Yes they both require you to factor the equation

- anonymous

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

- wolf1728

Okay then if you think you know it, then I am glad I helped you out.
Do you have any problems with part C?

- anonymous

im having trouble finding the gcf

- anonymous

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

- wolf1728

As for Part C I don't think that the equations have a common factor.
Unless they mean both equations have an x^2 in there and the factor of x^2 is 1.

- anonymous

thats the part i was mostly confused on.

- wolf1728

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

- anonymous

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

- UsukiDoll

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

- UsukiDoll

for A
(x +2) * (x +3)
for B
(x -2) * (x -2)
unless the signs were swapped in B, there isn't a common factor.

- UsukiDoll

and congrats @wolf1728 for reaching 95SS :)

- wolf1728

Thanks UsukiDoll :-)

- anonymous

Thank you both.

- UsukiDoll

you're welcome :)

- wolf1728

u r welcome

- anonymous

can you help me on 3 more questions?

- wolf1728

Three more??? Well okay. :-)

- anonymous

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

- anonymous

A function is shown below:
f(x) = x3 + 2x2 - x - 2
Part A: What are the factors of f(x)? Show your work. (3 points)
Part B: What are the zeros of f(x)? Show your work. (2 points)
Part C: What are the steps you would follow to graph f(x)? Describe the end behavior of the graph of f(x). (5 points)

- anonymous

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

- wolf1728

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

- wolf1728

You are now factoring a cubic equation

- anonymous

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

- anonymous

totally**

- UsukiDoll

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

- anonymous

im new that this.. @UsukiDoll

- wolf1728

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

- UsukiDoll

sorry x.x

- anonymous

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

- UsukiDoll

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

- UsukiDoll

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

- anonymous

how did you get that

- UsukiDoll

I was talking to Shalante earlier...

- anonymous

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

- UsukiDoll

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

- anonymous

okay, thank you for your help!

- wolf1728

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

- UsukiDoll

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

- wolf1728

You mentioned grouping

- UsukiDoll

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

- anonymous

no one is picking on you...

- wolf1728

I found a method for factoring a cubic
http://www.wikihow.com/Factor-a-Cubic-Polynomial
it is a 12 step process!

- UsukiDoll

oh no way... why do that when for the first two terms we can take out a x^2 and for the last two terms, take out the negative sign

- UsukiDoll

\[\large f(x) = x^3+2x^2-x-2 \]
on the first two terms \[\large x^3+2x^2 \] we noticed that there is a x^2 in common. we can see this easier by expanding. the exponent 3 means write x 3 times. similarly the exponent 2 means write x 2 times
\[\large xxx+2xx \]
we can yank out 2 x's
\[\large xx(x+2) \]
\[\large x^2( x+2) \]

- UsukiDoll

so for the second half we just factor out the negative
\[\large f(x) = x^3+2x^2-x-2 \]
\[\large -x-2 \]
\[\large -(x+2) \]

- anonymous

@UsukiDoll
It tells us to write in squares
(x-2)^2 is a square.
This is new in algebra
It like basic math telling us to write the solution of 5/8 +10/8 in mixed number form instead of reciprocal terms.

- UsukiDoll

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

- UsukiDoll

@Shalante can you stop interrupting me please?

- anonymous

part b is worth 3 points. guess he wont get all the credit. I would have replied earlier, but my laptop went ham

- UsukiDoll

anyway, we're still not done with the factor by grouping. We have one common term to yank out
\[\large f(x) = x^2(x+2)-(x+2)\]
so we notice that there is a (x+2) in common
uh oh. I forgot about one thing.
there is usually a 1 written in front of the -(x+2)
\[\large f(x) = x^2(x+2)-1(x+2)\]
so now yanking out the (x+2)
\[\large f(x) = (x+2)(x^2-1) \]

- UsukiDoll

we can factor the \[\large x^2-1\] using difference of squares
\[(a^2-b^2) = (a+b)(a-b)\]
letting a = x and b =1
\[(x^2-1^2) = (x+1)(x-1)\]

- UsukiDoll

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

- anonymous

thank you so much for explaining this to me.

- UsukiDoll

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

- anonymous

now what about the steps you would follow to graph f(x)? and Describe the end behavior of the graph of f(x)

- UsukiDoll

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

- anonymous

2,1,1????

- UsukiDoll

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

- anonymous

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

- UsukiDoll

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

- wolf1728

-2 -1 and1

- UsukiDoll

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

- UsukiDoll

I overlooked by accident.

- wolf1728

you did the factoring though

- anonymous

made the same mistake.

- UsukiDoll

It was an accident... geez.

- wolf1728

but you factored it correctly Usuki :-)

- anonymous

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

- wolf1728

well I'm gonna get going

- UsukiDoll

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

- anonymous

nobody is nitpicking, your helping me out alot

- UsukiDoll

tell that to @Shalante then. I'm through with this.

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