Homework Help

[Angel]

Use this topic to get help on HW questions or topics that you're confused about.
I guess you could consider this an FWG Tutoring Topic *shrug*

Math 1050:

The answers are -1,-2,1,2
Why are the answers both positive and negative and how to I find the negative answers???

[BIG EYES]

x^4-5x^2+4

To reduce the formula, you can utilize factors of what C is.

In this question, C = 4, the factors of 4 are 1, 2, and 4. Any of these numbers could be used, but two of these numbers when added together will equal B which is 5.

1 and 4 are factors that fulfill the requirements for the reduced equations.

Next going onto breaking down the equation, You can breakdown the middle term into two factors that equal up to -5x^2. Going back to the first part about factors, we know that -1 and -4 when added together = -5. Because of this -5x^2 can be rewritten as -4x^2 + -x^2.

The equation is now x^4 - 4x^2 - x^2 + 4, Now we can divide this entire equation by placing parentheses between the first two numbers and the last two numbers. Now it is (x^4 - 4x^2) + (-x^2 + 4).
We can reduce the first part of the equation by taking the variables x^2 out of the first parentheses. It will now become x^2(x^2 - 4) + (-x^2 + 4).

onto the reason why you have 4 answers

x^2 - 4 = 0
x = 2 ; (2^2 - 4) = 0 ; (4-4)= 0
x = -2 ; (-2^2 - 4) = 0 ; (4-4) = 0
x^2 - 1 = 0
x = 1 ; (1^2 - 1) = 0 ; (1-1) = 0
x = -1 ; (-1^2 - 1) = 0 ; (1-1) = 0

http://www.tiger-algebra.com/drill/x~4-5x~2_4=0/#skipad
This website also explains it

[Boxorino]

[Angel]

Dang... thanks you guys!

But... Does anyone know how to do it with U-Substitution and/or Synthetic Division?

[Lamb]

Dang... thanks you guys!

But... Does anyone know how to do it with U-Substitution and/or Synthetic Division?

U substitution O-o... I only do that in Calculus, I can look it up for algebra (If there is, I might be wrong and this is straight integration that you're looking for) and explain it though.
Synthetic division.. I've done it in precalculus, but I don't recall how to do it exactly, because it's algebra 2 iirc. I will look through my notes from last year to see if I have work that I can use to explain it.

[Lamb]

Ok, well I'll try my best to explain the two in words.
Let's begin with Synthetic Division. This is a short hand form of dividing polynomials.
Generally, you do this when you have a binomial as your divisor and a trinomial as your dividend.
Ex.
x-4 is your divisor
x^2-8x+16 is your dividend
What you would do with your divisor is set it equal to zero, and solve for x. x-4=0. x=4
You would then take the leading coefficients of your dividend and place them in order inside of a bracket that looks like an upside-down division bracket.
It should look similar to this
4 | 1 -8 16
__|______

What you do then is bring down the leading coefficient.
4 | 1 -8 16
__|______
----1
Then multiply that by your divisor and bring it to the top below your next coefficient.
4 | 1 -8 16
__|___4___
----1
Then you add the coefficient and the product and place it at the bottom of the bracket.
4 | 1 -8 16
__|___4___
----1 -4
Lastly, you repeat the same step again as before with multiplication, and add
4 | 1 -8 16
__|___4-16_
----1 -4 0
You are left with the coefficients to the other factor
(1)x+(-4)

This is just a faster way of doing division for polynomials, and is only applicable, if I remember correctly, when you have a factor that goes evenly into the polynomial, resulting in another factor. Think of it as a short way for Long division.

[Lamb]

U- Substitution is an integration technique.
I'll begin with an easy example and move on to a harder one.
Because the forum doesn't support the integral symbol, I will use int(
int(e^x dx)
we will substitute our x with u.(hue hue) u=x
we can then get dx for du. d/dx u= du=1dx
int( e^u du)
d/dx e^u = e^u du
so we can just do it. ( JUST DO IT)
int(e^u du) = e^u +C
we have that plus C for anything that may have resulted in a zero when we derived. Hue Hue.

Now let's get a little harder. No sexual innuendos pls

int( (x^2+4)/(x+2) dx)
We can substitute x+2 for u, and derive our other equations from this.
u=x+2
x=u-2
du=1dx
NOW SUBSTITUTE
int(((u-2)^2 +4)/(u))
now we have to foil out
int( (u^2 + (-4u) +4) +4)/(u))
then we can split it and pull out coefficients THE INNUENDOS ARE STRONG
int(((u^2)/u)du) + (-4int((u/u) du)) + 8int(du/u)
We can then simplify and integrate
int(udu) -4int(du) +8int(du/u)
(u^2)/2 +(-4u)+8ln|u|+C
Then we just substitute x+2 back in for u.
((x+2)^2)/2 -4(x+2) + 8ln|x+2| + C

Did you C your mistake? (a good way to remember that plus C.)
We could simplify our answer and pull out all the integers and eliminate them with the plus C, but personally I would stop here.
This stuff is purely practice q-q... Even now I'm still grinding through this.

[Angel]

OK!
So I asked the teacher to go over the question in class this morning so I could see how she wanted us to do it...and come to find out...I had done it right, but I had forgotten that when you take the square root of something, you have to make it plus or minus ("Plusorminus" as my professor calls it because "it's a nasty little dinosaur")

[Lamb]

OK!
So I asked the teacher to go over the question in class this morning so I could see how she wanted us to do it...and come to find out...I had done it right, but I had forgotten that when you take the square root of something, you have to make it plus or minus ("Plusorminus" as my professor calls it because "it's a nasty little dinosaur")

I did that when I first took pre-cal.
Skipped algebra two so I did not do |x|= +-n when I had to. (all the even functions) o-o