Well this week was actually better than the first two. I'm starting to get use to how b-rob teaches. One thing I understood from chapter three was finding inequalites. Sloving inequalites is basically just like solving for x. You bring the constant over to the side with the greater than or less than sign, and then solve for x. Make two equations with the one they give you, on the second equation change both signs. The only time you will use and/or are with absolute value. And if have to change the signs when you divide or mulitply by a negative number. Here are some examples that help explain all of this.

3x-9>4
3x-9<-4 3x-9>4 <----These are the two equations you will make. Notice how I changed the
3x<5>13 sign on the second one.
x<5/3>13/3

-7x-12<9
-7x<21
x>3 <----Since there was a negative to divide by the sign changed in the final answer.

One thing I didn't understand about inequalities was when theres an equation with fractions.
Don't you have to find the CGF? but I don't understand what to do after that. I think you have to multiply that by the fraction on both signs of the sign. I don't know, I'm confused. Someone explain please.

Reflection #3

okay, so this week went by really quick, but i think i caught on pretty quickly. The easiest thing for me was the inequalities.



3x-1>2



You just add one to both sides...3x>3... and then you divide by three on both sides...x>1... and that is your answer. But you have to remeber if you multiply or divide by a negative number, then you have to switch the signs. For example,



-2x+1>3



You would subtract one from both sides...-2x>2 and then you would divide by negative two...x<1...>

The absolute zero ones were easy too. If you have on greater than or less than, it is an OR, but if you have one that is greater than or equal to or less than or equal to, it is an AND. For example.

(!) is the absolute value sign becuase i can't find nothing else that looks like a line.

! x-4 ! <5

So you make two problems:

x-4<5>-5

x<9>-1

Example of the AND problem:

! x-9 ! >or equal to 2

Instead of making two problems, you just add it to the front of the original problem:

-2>or=to x-9 >or=to 2

Then you add nine to all sides...7>or=to x >or=to 11

So your final answer would be x<or=to 7 AND x>or=to 11

We also went over how to solve for x again and something clicked and i get everything now. I know how to solve by grouping, quadratic form, and rational root theorum. I new how to do everything on the test and i hope that i got an A.

But the hardest thing for me was on Friday when we learned how to do the domain and range. I hope it was just hard for me just because we just learned it and not because i'm stupid, but if yoyu get the domain and range thing, just tell me because i am so confused. Hopefully we will do more problems on Tuesday.

Reflection #3

Alright week three no different from any other week yet this year, we learned chapter 3 and reviewed chapter 2. we learned how to solve inequalities and find the domaine and range. The thing I understand the most in chapter three is how to solve inequalities for example I will solve one.
l3x-9l>4
Step 1:is it and or or
or
Step 2: when solving and there is an absolute value you must have two equations
3x-9<-4 or 3x-9>4
Step 3:solve for x.
x<5/3>13/3
**but when you divide by a negitive you must change your inequaitie sign**
one thing i didn't seem to get a hold of today was the domain and range. i don't understand how you get a number? i also don't understand when you use infinity.

heyyyy

i hav a question how would you do this prob.

4-2i

o and what page is our test review on

I have a big and important question, i f i dont get a right answer from somebody i'm gonna go buck wild!!!!!!!!!!!!!!!!!!!!!
i'm gonna give a problem and yall can solve and give me step by step details:)
find the roots and the equation given only this 3x+6

so wait....

for inequalities, the only time u do a and or a or is when u have absolute values?

Inequalities question?

hey if anyone can help me out with problem number 10 on page 98 that would be great.
here's the problem:

4/3 (x - 1/2) + 1/2x > 2/3 (2x - 5/2)

**and by the way, thats ^^supposed to be greater than or equal to. i don't know how to type that. hah

Well anyway, the first thing I did was distribute the 4/3 and the 2/3 but then I don't know what to do with all the fractions after that. I get kinda confused so some help would be good :)

anyone get this?

number 5 on page 98.

1/4(x-1) < or =" x+4/6

Reflection #2

Understand?
Well, I understand how to do the rational root theorem.
ex. x⁴ + 2x³ – 7x² – 8x + 12
factors of p(constant)=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
factors of q(leading coeff.)=+/-1
Find p/q=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
Plug all possibilities into table on calculator and find all where y is 0.
You should get 1, 2, -2, and -3.
Do cythetic division with any two.
1 1 2 -7 -8 12
1 3 -4 -12
1 3 -4 -12 0
(x^3+3x^2-4x-12)(x-1)
2 1 3 -4 -12
2 10 12
1 5 6 0
(x^2+5x+6)(x-1)(x-2)
Solve by factoring x^2+5x+6=
(x+2)(x+3)
x=-2, -3, 1, 2
Put in point form:
(-2, 0)(-3, 0)(1, 0)(2, 0)

Now for what I don't understand! Much easier.
I don't understand how to do the word problems. (Like the one problem with the barn and area and...??) I pay attention when she explains it in class, but I still don't know what I'm doing. Could anyone explain?

Reflection #2

Ah, whatta week! Not only did i supposedly have swine flu and become quarantined from school for about 2 days, but i also missed a whole lot in advanced math! :/
I'm still trying to catch up actually....
So far, i'm really comfortable with synthetic division so let me try to explain this....

Synthetic division is strictly used to find zeroes or roots of polynomials.

Equation: (x^5+x^4+2x^3+5x^2-3x+6) / (x+3)
First you'll set x+3=0
x=-3

this gives you what to synthetically divide by(not sure if that's the correct way to say that! lol) :

-3|1 1 2 5 -3 6




1st: Bring your first number down.

-3|1 1 2 5 -3 6
------------------
1

2nd: multiply the number at the bottom by the number being divided and add.

-3|1 1 2 5 -3 6
| -3 6 -24 57 -162
---------------------------
1 -2 8 -19 54 -156

Remainder for this equation would be:
R= -156.

All in all,
these are the steps in words...

• Bring down the 1.
• Multiply it by the -3 to get -3.
• Add 1+-3 to get -2.
• Multiply -2 times -3 to get 6.
• Add 2 plus 6 to get 8.
• Multiply 8 times -3 to get -24.
• Add 5 plus -24 to get -19.
• Multiply -19 times -3 to get 57.
• Add -3 plus 57 to get 54.
• Multiply 54 times -3 to get -162.
• Add 6 plus -162 to get -156.

-----------------------------------------------------------------------------------------

Now what i don't understand:

I have no idea how to come up with an answer for the Rational root theorem.
I can't figure out my notes for the Calc commands either?

I'm confused after i find the multiples of each,
where do i go from there?(Rational Root theorem)

I have read, reread, and reread again all of my notes. (including the insane amount of sidenotes i took from discussion) and i am still completely lost.
Of course, missing school doesn't help either but i'm really uncomfortable and if anyone has the EXACT calc commands please let me know.
here's what i wrote in my notes:

Y=
enter equation.
2nd mode(quit)
2nd, graph
2nd ZOOM?
2nd TRACE?!

I don't exactly know where i'm getting all of this from?
am i right? wrong? insane? lol.

any form of advice would be beneficial to me,
please&thanks! :)

This week in Adv Math the week went really fast and we had a lot of problems and equations to keep up with. I find that I picked up on more things than the first week though. After not doing very good on the first test I think I did a little better this time. These hundred and fifty dollar calculators are also coming in handy. One concept I did learn exceptionally well is Sketching Polynomials with out the finding the max. and min.

1st make sure the problem is factored all the way.
y = (x+1)(x-2)(x-4)

2nd take the roots and make a number line with a number between each root.
<----1---0---+2----+4--->

3rd find out for the sketching the positives and negatives
x(-2)neg,neg,neg= Negative
x(0)pos,neg,neg= Positive
x(3)pos,pos,neg= Negative
x(5)pos,pos,pos= Positive

4th put the original points on a graph
Graph -1, 2, and 4


5th sketch the graph by using the positives and negatives to know which way the lines should go
It should go negative, positive, negative, positive.

***********************************************************************
I know how to do the sketching on paper and on the calculator, but I do not know of the buttons to press on the calculator that will give me the max and min. Because I rember something about going left and right on the graph after you get to the max and min screen. Can somebody help me?

reflection 2

... the only thing i caught onto this week was synthetic division. that is probably the only thing in adv. math i'll understand, so let me explain it. say you are given a question find the remainder of (x+2)(3x^3+x^2-3):

step 1. make yourself a little box and stick -2 in it ( it's -2 bec. when you set x+2=0 that is your answer)

step 2. take all of your coefficents and put the on the side in a line.

( the red number is in the box) notice i have a zero between 1 and -3, that is because even though my x^1 isn't there it is still equal to zero

-2 3 1 0 -3

step 3. bring down your 3 and multiply it by -2 your answer is -6 after that take the -6 and put it right under the 1 and add giving you -5. That then drops down and you multiply -5 by -2 giving you 10. repeat thru -3

-2 3 1 0 -3
-6 10 -20
3 -5 10 -23

step 4. finally the remainder is -23.

if anyone can help with explaining when to use factor by grouping, or the root theorem that would be helpful. thanks.

Reflection #2

Okay, so i understood chapter 2 a little bit more than chapter one. The one thing that i thought was the easiest was the rational root theorum. Even though it is the longest step to do, i can do it faster than all of the other steps.

2x^4+x^2+7x+1=0

You press the Y= button on you calculator and type the eqation in. (make sure that you type everything in right)

First, you take the roots of the constant (p) and then the leading coefficient (q).
You would get p= +-1 then you would have q= +-2 +-1. Then you have to make p/q.

Possible roots: 1/2, -1/2, 1, -1

Then you plug all of your roots into your calculator. (hit 2nd, Graph) and it brings you to the table. Whatever number(s) you plug in and you get zero, use those numbers.

Next, you use synthetic division. You use the number(s) in the table that gave you zero.
Once you use synthetic divison, you should get all of the roots and your problem should be done.

___________________________________________________________________

Another thing that i learned this week was how to sketch a polynomial function.

y=(x+2)(x-1)(x+3)


1. The first step is to factor completely. (done)

2. The second step is to find your zeros and set up a number line.

x= -2 x=1 x=-3 <---(-3)(-2)-----(1)--->

3. The third step is to plug in any number on both sides of the roots.

f(-4)= -ve -ve -ve = -ve
f(-2.5)= -ve -ve +ve = +ve
f(-1)= +ve -ve +ve = -ve
f(2)= +ve +ve +ve = +ve


4. The fourth step is to sketch your graph. +ve are above the x axis, and -ve are below the x axis.

5. Plug in your calculator to see if your graph is correct.

6. You find your max. and min. in your calculator.

____________________________________________________________________

The thing that i thought was more difficult to learn was the other 2 was to solve a quadratic. I didn't really grasp the concept of how to do quadratic form and grouping. I kind of have an idea of how to do both of them, i just get lost in the middle of their steps. I wish that we would have spent more time on both of those methods, but other than that, chapter two was pretty easy.

Reflection #2

One thing i understood in chapter 2 was sketching a polynomial.

1. Your first step is to solve for y. Most problems that b.rob gave us are already solved for y.

y=(x+2)(x-4)(x+1)

2. Draw a number line and label your zeros

<--(neg 2)-(neg 1)---(pos 4)-->

3. Test out each number to the left and right of the number on the number line by plugging it into your function.

4. you will get +ve or -ve which will help you out a lot when graphing it

5. set up a graph and follow your +ve and -ve

(+ve means above the x-axis and -ve means below the x-axis)

6. once you graph the equation...plug the equation into your calculator and hit graph

this will allow you to see if your work is correct...

7. once you've plugged it into your calcuator you can find the maxium or minium...THIS CAN ONLY BE ONE YOUR CALCUATOR!!

On the other hand, I'm 110% completely lost when it came to the word problems this week. I didn't really know the steps except for the us b(squard)/2a.......................

Reflection #2

Alright, week two just like week one flew by but the some of the things I caught on to very quickly was definitely using my calculator, how to find the sum of the roots, how to find the product of the roots, maximum, minimum, and how to sketch a polynomial function. An example of how you sketch a polynomial, solve for the maximum, and minimum:
Given this equation: y=(x+1)(x-1)(x-2)
Step 1: factor the problem completely-done
Step 2: graph the completely factored equation on a number line and lable your zeros
Step 3: plug in numbers on either side of you zeros
Step 4: +ve-above x-axis
-ve- below the x-axis
Step 5: plug in your original equation into y= in your calculator and check your graph.
To find the maximum and minimum:
Step 6: you can only use your calculator
Quadratic maximum and minimum
Step one: x=-b/2a the vertes is always your maximum or minimum

Something I had trouble learning is how to solve anything bigger than a quadratic:
I don’t under stand how you start off with x^4-7x^2-8=0 and get g^2-8g-8=0?
How does the equation start off with the leading coefficient a number raised to the 4th but ends up being a number raised to the 2nd?

Reflection 2

This week I thought i learned alot, but I made little mistakes and that messed me over. The thing I actually learned about was graphing polynomials. I stayed up practicing that for an hour, AND THERE WAS ONE QUESTION ON THE TEST WITH IT!!! The easiest thing I think we had to do in any of our lessons this week, step wise, was synthetic division. I actually understand that. All you have to do is take the number in front of the x's and put them in a list. After that you bring the first number down and multiply it by the number your dividing by then add or subtract the numbers. Sorry if nobody understands what I just said, but you all should know what I'm talking about and if you don't then you fail just like me:) Something I did not remember would def. that 2+i and 2-i stuff tho. I also learned this week that if you are given the screen of the calculators graph then you can give the equation just by looking at the number the graph goes through, changing the sign, then put it next to x.

Reflection 2

learning chapter 2 was difficult, I had some trouble with quadratic form and rational root theorem, but I did learn how to use the calculator with synthetic division, graphing, and checking answers, and I have to say the easiest section was sketching polynomials

Sketching Polynomials:

1) Factor completely

2) Set up a number line and label zeros

3) Plug in numbers on either side of your roots

4) +ve=above the x axis
-ve=below the x axis

5) use the calculator to check results

6) find the max. and min. in calculator only

ex: y = (x+1) (x-1) (x-2)

1)done
2) -1,1,2
3)f(0)= +ve
f(1 1/2)=-ve
f(3)=+ve

max-(-.215 , 2.113)
min-(1.549 , -.631)

Rational Root Theorem:

1) Find all possible roots p/q p=factors of the constant
q=factors of the leading coefficient

2) Check to see which roots work in the table

3) Use synthetic division to factor the working roots

4) Solve the quadratic

Ex: x^3 +5x^2 -4x - 20=0

(x^3 + 5x^2) - (4x + 20)
(x^2 - 4) (x + 5)
x= -4 , +-2 , -5

reflection #2

One of the things we learned this week was how to sketch polynomial functions.

y=(x+1)(x-1)(x-2)

1. factor completely

the problem is already factored out

2. set up a number line, and label zeros

<=-2=-1===1=>

-2, -1, 1

3. plug in on either side of your roots

f(-2) (x+1)(x-1)(x-2)= -ve

f(0) +ve -ve -ve = +ve

f(1.5) +ve +ve -ve = -ve

f(3) +ve +ve +ve = +ve

4. +ve above x-axis; -ve below x-axis

5. check in calc.

6. max and min - in calc only

quadratic max and min

x=-b/2a (vertex is always max or min)

Reflection #2

Ch. 2 like ch. 1 was difficult, but I grasped the sketching a polynomial the best I think.

How to sketch the graph of a polynomial function:

1) Make sure the equation is factored completely. This may be done for you already but if it is'nt you MUST factor or the process won't work.

2) Draw a simple number line and on it label your zeros(roots, x ints., its all the same)

3) Now what you want to do is take a number on both the positive and negative side of your roots on the number line, and plug them into the polynomial function. When you do this, you do not have to actually multiply the number. The important thing is to keep track of negatives and positives. For example:
Say you need to plug 2 into (x-5)(x+2)(x-1)
Instead of worrying about adding and subtracting, you would plug it in and write down this:
-ve x +ve x +ve = -ve

4) Sketch a graph by placing your x-intercepts on a graph, then follow what you plugged in (previous step) to see where you will be above or below the x axis:
+ve = up
-ve= down

5) Plug the equation in the y= on your calculator and press graph to check yourself, if you graph looks similar to the graph on your calculator, you're good!

6) Find Minimum and Maximum in your calculator.

___________________________________________________________________
My only problem this week with not understanding something was with word problems. It was'nt that I didn't know the steps to follow but I get to a certain point where I have the information I need and I just can't figure out what I need to do with this information. The more I see them done the better it is getting though so I think the answer is just keep practicing them and applying what i know.

Reflection#2

And another great week has gone by at riverside academy. This week was a little bit tougher in advance math because i was never exposed to solving problems that way, but for the most part i though i keep up at a pretty good pace. I also learned this week that i can't do enough practice work and homework. Just when i think i get it mrs. robinson goes and throws another curve ball at me, but thats fine with me becuase i need a teacher that will push me and make me work hard. Anyways this week in math we learned the three different ways you can solve a a problem that is bigger that a quadratic. We learned how to do the grouping way, we learned how to do it by using the g way, and we learned how to do it by doing the rational root theorem. Using the rational root theorem definitly takes the longest amount of time, i also found that it was one of the most confusing and hardest way to solve a problem. The hardest stuff i thought we did was the area problems. I guess i was so use to easier area problems that this stuff was like greek to me, but i thought i was getting the hang of it. The easiet of the problems was doing it by grouping. It was so simple. If you have a problem like 5xto the 4th plus 5xto the third plus 2xto the second plus 2x you group the problem and take out whats alike. It should leave you with (5xto the third plus 2x)+(x+1) than you simply solve for x. I did think the test this week was super hard though. I spent half my studing time in groups and i still had trouble with the test. I guess we need to work harder problems, but i mite need some explaining on the area problems on the test

Reflection 2

First of all i would like to start off saying that Chapter 2 was definitely difficult to learn. There were some lessons I was able to understand very easily, but the others took a lot of practice. To prepare for the Chapter 2 test there was plenty of studying involved. I had to read over notes from Chapter 1, as for the same in Chapter 2. One of the lessons I seemed to understand the most was the Rational Root Theorem.

Rational Root Theorem:
1.) Find all possible roots where p is all factors of the constant; q is the factors of the leading
coefficient; p/q.
x^3-x^2-x+1
p=+ or - 1
q=+ or - 1
p/q= 1, -1

2.) Check to see which roots work in the table.
Calculator steps:
+ Click y=
+ Plug the problem into calculator.
+ Click 2nd mode(quit).
+ Click 2nd graph.
+ Plug in all possibilities in the x column.
+ Once possibilities are found, click 2nd mode(quit) again.

3.) Do synthetic division to factor all roots that work.
1 1 -1 -1 1
1 0 -1
_________
1 0 -1 0

4.) Solve the quadratic.
(x-1)=0 (x^2-1)=0
x=1 x^2=1
x= + or - 1
(1,0) (1,0) (-1,0)

I then seemed to find that the most trouble I had with Chapter 2 was the word problems. I understood the formulas for Area and Volume: A= l X w V= l X w X h but I still had trouble working out these problems.

Overall, Chapter 2 was a hard chapter to grasp and hopefully I will understand the next chapter a little better.

Reflections 2

Well this past week was a little more difficult than last week. There was more material to learn and I didn’t study as much as I should have. I knew everything when we were going over it in class however, when it came to the test my mind went blank. Here are a few of the things that gave me trouble on the test: Finding a quadratic when given the roots. Say the roots were

Step 1: find the sum of the roots:

Step 2: find the product of the roots: (

Step 3: Write the equation: x^2 – (sum) x +product=0

Therefore, the equation will be:

The sum of the roots of a polynomial = -2^nd coefficient/leading coefficient

The product of the roots of a polynomial are: If the degree is even= constant/leading coefficient

: If degree is odd = -constant/leading coefficient

For example: 6x^3-9x^2+X=0

The sum of the roots= 9/6= 3/2

The product of the roots= -0/6=0

Those were the only problems I remember problems with on the test. I’m sure I had problems with a few other things, but all I have to do is study more and the next test should be a piece of cake. Overall there was just a lot of information to learn and not very long to learn it. Once we start getting into the year I think things will get better. Thanks guys, hope my reflection helps someone out. -Evin

Reflections # 2

Okay, this week went by at about a medium pace. I had very little trouble understanding everything, and it was pretty cool learning all the new methods for solving for x. I mostly understood the rational root theorem, factoring by grouping, and the quadratic form. I am going to specifically explain the rational root theorem.

Rational Root Theorem:
Example problem

3x^3 - 5x^2 + 5x - 2 = 0

You must find p/q. P is the factors of the constant, and q is the factors of the leading coefficient.

P (factors of -2) = +-2, +-1
Q (factors of 3) = +-3, +-1
p/q = +-2/3, +-2, +-1/3, +-1

Now go to y = in your calculator and plug in the equation. Hit 2nd Graph, and plug in the p/q's to find the right possibility. Once a factor = 0, it is correct, and you can use it.

So if you plug in 2/3, your y should = 0.
Now you use synthetic division and divide the equation by 2/3.

2/3

3 -5 5 -2
2 -2 2
3 -3 3 0

So the new equation is 3x^2 - 3x + 3 = 0
Divide everything by 3. Then use the quadratic formula to solve for the remaining roots.
x^2 - x + 1

x = -b +- the square root of b^2 - 4ac all over 2a
x = 1 +- sqrt of 1 - 4 (1)(1) all over 2 (1)
x = 1 +- sqrt of -3 all over 2
x = 1 +- (i) sqrt of 3 all over 2
x = 1/2 +- i sqrt of 3 over 2

So the final roots are (1/2 + (i) sqrt of 3 over 2, 0)
(1/2 - (i) sqrt of 3 over 2, 0)
(2/3, 0)

reflection 2

Soo, yeah I'm like a week behind, ha..but whatever. I thought this week went by extremely slow. I learned a lot of things that I didn't think I would actually get. Like the 3 choices to solve anything bigger than a quadratic. I think the Rational Root Theorem is the easiest because synthetic division is so easy. Especially since you can just plug the equation into your calculator and put the possibilities in the table. I don't really get the quadratic form one though, when you make g equal to x. I know you take the exponent of the first term and put it over 2. Then you put g in for x. After that I am so confused. Well, not really i know you have to factor but somewhere during the factoring i get confused and don't get the same answer as everyone else.

One thing that I understand and can teach people how to do it if they don't understand is to sketch polynomial funtions.

If you have y=(x+1)(x-1)(x-2)

1. check if its factored completly
2. set up a number line..
-1, 1, 2
<-------------------------->
-1 1 2
3. f(-2)- neg-neg-neg=neg
f(0)- pos-neg-neg=pos
f(1.5)- pos-pos-neg=neg
f(3)-pos-pos-pos=pos


4. then draw your graph, can't really do that on here..ha

5. plug your equation into your calculator and check to see if your graph is correct

6. then find the max and min, which again..can't do that on here..



If anyone can help me with the quadratic form g thing let me know:)

Reflections 2

This week was a little slow. Although the math was a little harder this week, it still sunk in and dug deep into my brain. Let's hope that it stays there for the future. I mean, we have some genii in the class, and then theres the few slow people that get it, but it takes some time to get it thoroughly. Let's see, the homework this week was on overkill, but it all came together in the end. I gotta say, my new way of studying helped. I got this program called skype where you can call other people that are online, or you can text chat with them, and you can video chat too. Chase, Justin, Stephen, and me all used skype to study with each other on Wednesday and Thursday night, and it really did help. All you have to do is have a microphone, or not, to use skype. We studied for a while, and it was way easier studying with a group of people instead of myself. It's actually more convenient than going all the way to other people's houses to study with a group, and it's free, except if you want to call actual phones, then you have to pay, but it's better just to use computer calls. If you want to download it and check it out, go to skype.com and download it, contact me if you get it and need some help figurin out how to use it and all that jazz.
So, back to math.

One thing that i did understand easily this week was the part of the chapter where it said factor by grouping.

ok, so take x^3 + 5x^2 - 4x - 20 = 0

to do this problem, first, group the terms (x^3 + 5x^2) - (4x - 20) = 0

then, reduce the terms (or whatever it's called) so that x^2 (x + 5) - 4 (x + 5) = 0

then, regroup the terms outside of the parenthesis so that (x^2 - 4) (x + 5) = 0

then, take the two groups and solve for x x^2 - 4 = 0 and x + 5 = 0
x^2 = 4 and x = -5
x = ±2

then, put your answers in point form (-2,0) (2,0) (-5,0)

then, you're done


But one thing that i didnt get too quickly and need a little bit more help with is problems that you have to find area and you have to draw a box most of the time and all that stuff. I really get confused with that kind of stuff.

Reflection #2

This week wasn't too bad. I had some trouble with finding out the equations of the sketch of a graph but other than that, no problems. The thing that i really understood was quadratic form.

Quadratic form works in a trinomial equation where the leading exponent is divisible by 2.

Lets say the problem is x^4+6x^2+8.

You would start by finding x/2 or g. In this case, g=x^2.

After fing g, you plug it into the equation. It should now look like this: g^2+6g+8.

At this point you can easily factor the equation and it will look like this: g(g+2)+4(g+2)

It will then come to (g+2)(g+4) which means g=-2, g=-4

And since we're solving for x and g=x^2, you have to find the square root of g.

The end result should come to: (i square root of 2, 0)(-i square root of 2, 0)(2, 0)(-2, 0)