Question

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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

but how are you getting these...

Answer 5

I guess you don't know how to factor.

Answer 6

sadly no...

Answer 7

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?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

so its just like part a?

Answer 13

Yes they both require you to factor the equation

Answer 14

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

Answer 15

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

Answer 16

im having trouble finding the gcf

Answer 17

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

Answer 18

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.

Answer 19

thats the part i was mostly confused on.

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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.

Answer 24

and congrats @wolf1728 for reaching 95SS :)

Answer 25

Thanks UsukiDoll :-)

Answer 26

Thank you both.

Answer 27

you're welcome :)

Answer 28

u r welcome

Answer 29

can you help me on 3 more questions?

Answer 30

Three more??? Well okay. :-)

Answer 31

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

Answer 32

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)

Answer 33

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

Answer 34

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

Answer 35

You are now factoring a cubic equation

Answer 36

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

Answer 37

totally**

Answer 38

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

Answer 39

im new that this.. @UsukiDoll

Answer 40

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

Answer 41

sorry x.x

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

how did you get that

Answer 46

I was talking to Shalante earlier...

Answer 47

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

Answer 48

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

Answer 49

okay, thank you for your help!

Answer 50

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

Answer 51

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

Answer 52

You mentioned grouping

Answer 53

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

Answer 54

no one is picking on you...

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

so for the second half we just factor out the negative

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

Answer 59

@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.

Answer 60

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

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

Answer 61

@Shalante can you stop interrupting me please?

Answer 62

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

Answer 63

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)
\[\large f(x) = (x+2)(x^2-1) \]

Answer 64

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
\[(x^2-1^2) = (x+1)(x-1)\]

Answer 65

\[\large f(x) = (x+2)(x+1)(x-1) \]

that's the final factored form

Answer 66

thank you so much for explaining this to me.

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

2,1,1????

Answer 71

close....
solve for x

x+1= 0
subtract 1 on both sides

Answer 72

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

Answer 73

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

Answer 74

-2 -1 and1

Answer 75

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

Answer 76

I overlooked by accident.

Answer 77

you did the factoring though

Answer 78

made the same mistake.

Answer 79

It was an accident... geez.

Answer 80

but you factored it correctly Usuki :-)

Answer 81

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

Answer 82

well I'm gonna get going

Answer 83

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

Answer 84

nobody is nitpicking, your helping me out alot

Answer 85

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