calc homework that is actually algebra... factoring

[jmsic3345]

yeah so the first day of school my calc teacher gave us homework and i dont remember anything... so please help me ><

problem: factor and simplify. Express the answer as a fraction without negative exponents.

-5/2x(x+3)[sup:1niovcam]-3/2[/sup:1niovcam]+5(x+3)[sup:1niovcam]-1/2[/sup:1niovcam]
erm i think im supposed to factor by using the 5(x+3)[sup:1niovcam]-1/2[/sup:1niovcam]?
5(x+3)[sup:1niovcam]-1/2[/sup:1niovcam] * (1-1/2x[sup:1niovcam]-2/2[/sup:1niovcam])

and another one is

Express as a simple fraction.

( (-4/x+h)-(-4/x) ) / h
i thinkkkkk i would have to work out the h first which would make the equation be:
-4/xh+h[sup:1niovcam]2[/sup:1niovcam] + 4/xh?

maybe im completely wrong.... lol

 

[Deleted member 4993]

yeah so the first day of school my calc teacher gave us homework and i dont remember anything... so please help me ><

problem: factor and simplify. Express the answer as a fraction without negative exponents.

-5/2x(x+3)[sup:14cs9m4u]-3/2[/sup:14cs9m4u]+5(x+3)[sup:14cs9m4u]-1/2[/sup:14cs9m4u] <<< Among (-3/2) and (-1/2) - which one is smaller - you need to factor out smaller power.
erm i think im supposed to factor by using the 5(x+3)[sup:14cs9m4u]-1/2[/sup:14cs9m4u]?
5(x+3)[sup:14cs9m4u]-1/2[/sup:14cs9m4u] * (1-1/2x[sup:14cs9m4u]-2/2[/sup:14cs9m4u])

and another one is

Express as a simple fraction.

( (-4/(x+h)-(-4/x) ) / h
i thinkkkkk i would have to work out the h - No - not the way you are trying first which would make the equation be:
-4/xh+h[sup:14cs9m4u]2[/sup:14cs9m4u] + 4/xh?

first multiply the numerator and the denominator by x(x+h) to simplify the numerator

maybe im completely wrong.... lol

 

[jmsic3345]

ok so for the first problem i know im supposed to do:
5/2(x+3)[sup:yrtflynw]-3/2[/sup:yrtflynw][-1+......2(i think) but then i dont know what the exponent is supposed to be....

and for the second one i dont get what you mean... which numerators and which denominators because when i multiply by (x+h)x i get the exact same thing i started with... =.=
i think im stupid T_T

please helppp~

 

[mmm4444bot]

… for the first problem i know im supposed to do:

5/2(x+3)[sup:2yms1x9t]-3/2[/sup:2yms1x9t][-1 + …

Hello JM:

What happened to the other factor of x ?

You typed the first term of the original expression as follows.

-5/2x(x+3)^(-3/2)

This means the following.

-(5/2) * x * (x + 3)^(-3/2)

We won't get this from multiplying 5/2(x + 3)^(-3/2) by -1.

(Maybe, you mistyped your original post?)

Before I do any more work with your first exercise, I would like you to confirm that you properly typed the given expression. In other words, please confirm that it's the following.

\(\displaystyle -\frac{5}{2} \cdot x \cdot (x + 3)^{-3/2} + 5 \cdot (x + 3)^{-1/2}\)

On your second exercise, we need to simplify the numerator of the compound fraction, first.

This numerator is -4/(x + h) - (-4/x).

To simplify this expression, we need a common denominator before subtracting -4/x from -4/(x + h).

The common denominator is x(x + h).

Since you admit that you can't remember anything, I'll do the second exercise for you.

\(\displaystyle \frac{\frac{-4}{x + h} - \frac{-4}{x}}{h}\)

\(\displaystyle \frac{\frac{4}{x} - \frac{4}{x + h}}{h}\)

\(\displaystyle \frac{\frac{4}{x} \cdot \frac{x + h}{x + h} - \frac{4}{x + h} \cdot \frac{x}{x}}{h}\)

\(\displaystyle \frac{\frac{4(x + h)}{x(x + h)} - \frac{4(x)}{x(x + h)}}{h}\)

\(\displaystyle \frac{\frac{4(x + h) - 4x}{x(x + h)}}{h}\)

\(\displaystyle \frac{\frac{4x + 4h - 4x}{x(x + h)}}{h}\)

\(\displaystyle \frac{\frac{4h}{x(x + h)}}{h}\)

\(\displaystyle \frac{4h}{x(x + h)} \cdot \frac{1}{h}\)

\(\displaystyle \frac{4}{x(x + h)}\)

Let us know, if I wrote anything that you do not understand.

Please confirm the given expression in your first exercise.

Cheers ~ Mark

MY EDIT: Deleted comments about smaller exponent because I wasn't thinking clearly.

 

[jmsic3345]

uhhhh the equation you typed was right and i re wrote the equation wrong on my paper so yeah i didnt factor the x but my calc teacher said something about derivatives and limits or whatnot and that im supposed to factor the 5/2(x+3)[sup:3ai216c9]-3/2[/sup:3ai216c9] first because it is smaller...

 

[bayaron]

could some one help me write and equation for The line through (5,-3) and (2,-3)

 

[jmsic3345]

NUMBER 1)
to find the equation the formula you'd use (y[sub:k25iqx1n]2[/sub:k25iqx1n]-y[sub:k25iqx1n]1[/sub:k25iqx1n])/(x[sub:k25iqx1n]2[/sub:k25iqx1n]-x[sub:k25iqx1n]1[/sub:k25iqx1n])=m to find the slope and then use the formula y-y[sub:k25iqx1n]1[/sub:k25iqx1n]=m(x-x[sub:k25iqx1n]1[/sub:k25iqx1n]) and for the y[sub:k25iqx1n]1[/sub:k25iqx1n] and x[sub:k25iqx1n]1[/sub:k25iqx1n] you'd take any of the two points that are given and plug them in
NUMBER 2)
your problem should be in the beginning algebra section
NUMBER 3)
i has nothing to do with my problem... which doesn't help so yeah next time please put your problems elsewhere. ~thnx

 

[mmm4444bot]

… my calc teacher said something about derivatives and limits or whatnot and that im supposed to factor the 5/2(x+3)[sup:3vro1y3y]-3/2[/sup:3vro1y3y] first because it is smaller …

Ahh. Perhaps, I misspoke. I was thinking of factoring z^(-3/2) + z^(-1/2).

z^(-1/2) * (z^(-1) + 1)

But, we could also factor it like so.

z^(-3/2) * (1 + z)

Maybe, the second way is better, since there's no negative exponent inside the parentheses. :wink:

(My apologies to Subhotosh.)

Anyways, in the second factorization above, do you see why we need the z^1 term inside the parentheses ?

When that term is multiplied by z^(-3/2), the product needs to be z^(-1/2).

z^(-3/2) * z

z^(-3/2) * z^(2/2)

z^(-3/2 + 2/2)

z^([-3 + 2]/2)

z^(-1/2)

Actually, there's a bunch of different approaches. Here's another!

\(\displaystyle -\frac{5}{2} x (x + 3)^{-3/2} + 5 (x + 3)^{-1/2}\)

\(\displaystyle 5(x + 3)^{-1/2} - \frac{5x}{2} (x + 3)^{-3/2}\)

\(\displaystyle \frac{5}{2} \left( \frac{2}{(x + 3)^{1/2}} - \frac{x}{(x + 3)^{3/2}} \right)\)

\(\displaystyle \frac{5}{2} \left( \frac{2}{(x + 3)^{1/2}} \cdot \frac{(x + 3)^{2/2}}{(x + 3)^{2/2}} - \frac{x}{(x + 3)^{3/2}} \right)\)

\(\displaystyle \frac{5}{2} \left( \frac{2(x + 3)}{(x + 3)^{3/2}} - \frac{x}{(x + 3)^{3/2}} \right)\)

\(\displaystyle \frac{5}{2} \left( \frac{2(x + 3) - x}{(x + 3)^{3/2}} \right)\)

\(\displaystyle \frac{5}{2} \left( \frac{x + 6}{(x + 3)^{3/2}} \right)\)

\(\displaystyle \frac{5}{2} \cdot (x + 3)^{-3/2} \cdot (x + 6)\)

My final thoughts are that (1) factoring has nothing to do with derivatives, limits, or "whatnot", and (2) you might not be ready to begin studying calculus.

Do you still have your precalculus textbook? I suggest that you start some heavy review. Or, begin psychological preparations for self-torture.

When people are confronted with a lot of algebra and trigonometry that they don't understand, they can't really "see" the calculus.

Cheers

 

[Denis]

Re:

Actually, there's a bunch of different approaches. Here's another!
\(\displaystyle -\frac{5}{2} x (x + 3)^{-3/2} + 5 (x + 3)^{-1/2}\)

Or the way li'l ole me likes:
Let k = x+3 ; get common denominator:

[10k^(-1/2) - 5xk^(-3/2)] / 2

= 5k^(-3/2)(2k - x) / 2

= 5(2k - x) / [2k^(3/2)]

Roger?

 

[jmsic3345]

i talked to my calc teacher and i get it now... he said it was supposed to be factored:
5/2(x+3)[sup:2ivns7xr]-3/2[/sup:2ivns7xr][-x+2(x+3)]

 

[mmm4444bot]

… my calc teacher … said it was supposed to be factored:

5/2(x+3)[sup:3of95ois]-3/2[/sup:3of95ois][-x+2(x+3)]

From what you posted, in this thread, there's no way for anybody here to determine that required form.

More curious, to me, is why your teacher would not simplify -x + 2(x + 3). :?

 

[jmsic3345]

he did at the end... in the fraction form =.=
he wrote 5x+6/2(x+3)[sup:3gvyj46n]3/2[/sup:3gvyj46n] but i think he forgot the -x and instead added it... gahhh i hate this class aready T_T