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)

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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)

Mathematics
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Part A Basically isn't this just asking to factor the equation? (x +2) * (x +3)
yes but i have been having big problems trying to solve this things, i shouldn't be up crying over this....
Part B (x -2) * (x -2)

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but how are you getting these...
I guess you don't know how to factor.
sadly no...
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?
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
Okay then if you think you know it, then I am glad I helped you out. Do you have any problems with part C?
im having trouble finding the gcf
im new to this, its hard with no teacher or anyone to explain..
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.
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
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.
and congrats @wolf1728 for reaching 95SS :)
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...
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)
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...
I found a method for factoring a cubic http://www.wikihow.com/Factor-a-Cubic-Polynomial it is a 12 step process!
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
\[\large f(x) = x^3+2x^2-x-2 \] 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 \[\large xx(x+2) \] \[\large x^2( x+2) \]
so for the second half we just factor out the negative \[\large f(x) = x^3+2x^2-x-2 \] \[\large -x-2 \] \[\large -(x+2) \]
@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.
\[\large f(x) = x^3+2x^2-x-2 \] through factor by grouping we have \[\large x^2(x+2)-(x+2)\]
@Shalante can you stop interrupting me please?
part b is worth 3 points. guess he wont get all the credit. I would have replied earlier, but my laptop went ham
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) \[\large f(x) = (x+2)(x^2-1) \]
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 \[(x^2-1^2) = (x+1)(x-1)\]
\[\large f(x) = (x+2)(x+1)(x-1) \] 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
now what about the steps you would follow to graph f(x)? and Describe the end behavior of the graph of f(x)
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
tell that to @Shalante then. I'm through with this.

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