anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
wolf1728
  • wolf1728
Part A Basically isn't this just asking to factor the equation? (x +2) * (x +3)
anonymous
  • anonymous
yes but i have been having big problems trying to solve this things, i shouldn't be up crying over this....
wolf1728
  • wolf1728
Part B (x -2) * (x -2)

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anonymous
  • anonymous
but how are you getting these...
wolf1728
  • wolf1728
I guess you don't know how to factor.
anonymous
  • anonymous
sadly no...
wolf1728
  • 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
  • anonymous
so in this case, a= 2 b=5 and c=6 ???
wolf1728
  • wolf1728
If you are referring to part A then a would equal 1 and you have the rest correct.
anonymous
  • anonymous
sorry, i missed that.. i understand that part. thank you! but can you explain part B?
wolf1728
  • wolf1728
Okay for part B the equation is x2 - 4x + 4 and a =1 b=-4 and c=4
anonymous
  • anonymous
so its just like part a?
wolf1728
  • wolf1728
Yes they both require you to factor the equation
anonymous
  • anonymous
seems like i was stressing over nothing! somehow you made it clear to me, thank you so much
wolf1728
  • 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
  • anonymous
im having trouble finding the gcf
anonymous
  • anonymous
im new to this, its hard with no teacher or anyone to explain..
wolf1728
  • 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
  • anonymous
thats the part i was mostly confused on.
wolf1728
  • 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
  • anonymous
thank you, do you mind if you can walk me through some other ones?
UsukiDoll
  • UsukiDoll
I don't see a common factor for part c either
UsukiDoll
  • 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
  • UsukiDoll
and congrats @wolf1728 for reaching 95SS :)
wolf1728
  • wolf1728
Thanks UsukiDoll :-)
anonymous
  • anonymous
Thank you both.
UsukiDoll
  • UsukiDoll
you're welcome :)
wolf1728
  • wolf1728
u r welcome
anonymous
  • anonymous
can you help me on 3 more questions?
wolf1728
  • wolf1728
Three more??? Well okay. :-)
anonymous
  • anonymous
yes.. but this is the last time having to deal with this topic...
anonymous
  • 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
  • anonymous
for part a.... a=1 b=2 c=-2 not 100% sure...
wolf1728
  • wolf1728
Part A a=1 b=2 c=-1 d=-2
wolf1728
  • wolf1728
You are now factoring a cubic equation
anonymous
  • anonymous
totaly just misssed the "x" there.. sorry about that
anonymous
  • anonymous
totally**
UsukiDoll
  • UsukiDoll
factor by grouping for the first part rational root test for the second part
anonymous
  • anonymous
im new that this.. @UsukiDoll
wolf1728
  • wolf1728
and it's been a while since I factored a cubic equation
UsukiDoll
  • UsukiDoll
sorry x.x
anonymous
  • anonymous
B should be \[(x-2)^2 \] because that is in square of a linear expression formio
UsukiDoll
  • UsukiDoll
(x-2)^2 and (x-2)(x-2) mean the same thing.
UsukiDoll
  • UsukiDoll
For (x-2)^2 that exponent number is 2 meaning write (x-2) 2 times (x-2)(x-2) .
anonymous
  • anonymous
how did you get that
UsukiDoll
  • UsukiDoll
I was talking to Shalante earlier...
anonymous
  • anonymous
gotcha! i wish i could understand this, the first problem i got, but now this is totally different
UsukiDoll
  • UsukiDoll
I know... I got distracted... one sec while I cool off before going berserk.
anonymous
  • anonymous
okay, thank you for your help!
wolf1728
  • wolf1728
UsukiDoll it seems you know how to factor a cubic equation?
UsukiDoll
  • UsukiDoll
\[f(x) = x^3+2x^2-x-2\]
wolf1728
  • wolf1728
You mentioned grouping
UsukiDoll
  • UsukiDoll
I can't do anymore of this... I know it, but I'm being picked on.
anonymous
  • anonymous
no one is picking on you...
wolf1728
  • wolf1728
I found a method for factoring a cubic http://www.wikihow.com/Factor-a-Cubic-Polynomial it is a 12 step process!
UsukiDoll
  • 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
  • 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
  • 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
  • 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
  • UsukiDoll
\[\large f(x) = x^3+2x^2-x-2 \] through factor by grouping we have \[\large x^2(x+2)-(x+2)\]
UsukiDoll
  • UsukiDoll
@Shalante can you stop interrupting me please?
anonymous
  • 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
  • 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
  • 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
  • UsukiDoll
\[\large f(x) = (x+2)(x+1)(x-1) \] that's the final factored form
anonymous
  • anonymous
thank you so much for explaining this to me.
UsukiDoll
  • 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
  • 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
  • UsukiDoll
did you solve for x first in the previous part? there should be 3
anonymous
  • anonymous
2,1,1????
UsukiDoll
  • UsukiDoll
close.... solve for x x+1= 0 subtract 1 on both sides
anonymous
  • anonymous
some how im getting 2,1 and then -1 but im over thinking this and getting more confused
UsukiDoll
  • UsukiDoll
it's correct now x =2,1,-1
wolf1728
  • wolf1728
-2 -1 and1
UsukiDoll
  • UsukiDoll
oh wait I missed. ugh yeah wolf is right -2 -1 1
UsukiDoll
  • UsukiDoll
I overlooked by accident.
wolf1728
  • wolf1728
you did the factoring though
anonymous
  • anonymous
made the same mistake.
UsukiDoll
  • UsukiDoll
It was an accident... geez.
wolf1728
  • wolf1728
but you factored it correctly Usuki :-)
anonymous
  • anonymous
its late, im sure we all make mistakes.. its okay.. your doing a good job!
wolf1728
  • wolf1728
well I'm gonna get going
UsukiDoll
  • UsukiDoll
I'm out too. the nitpicking towards me is too much.
anonymous
  • anonymous
nobody is nitpicking, your helping me out alot
UsukiDoll
  • UsukiDoll
tell that to @Shalante then. I'm through with this.

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