# Thread: I am stuck again!

1. ## I am stuck again!

mathhelpforum users, I need your help!

While doing some homework, I got stuck on a few problems.

Find the simplest equivalent expression for each of the following:

4xsquared - 64/12xsquared - 60x + 48

2xsquared - 128/4xsquared - 36x + 32

Please show how you got the answer, so that I can learn from difficulties.
Any help will be appreciated!

2. Given: $\displaystyle 4x^2 - \frac{64}{12}x^2 - 60x + 48$

Multiply by 12 to get rid of the fraction: $\displaystyle 48x^2 - 64x^2 - 720x + 576$

Combine the $\displaystyle x^2$: $\displaystyle -16x^2 - 720x + 576$

You can pull -16 as common factor: $\displaystyle -16(x^2 + 45x - 36)$

Now do the same for question number 2

Do you also need to factor?

3. Hello, Joker37!

Please use parentheses when writing fractions . . .

(4xsquared - 64) / (12xsquared - 60x + 48)
We have: .$\displaystyle \frac{4x^2-64}{12x^2-60x+48}$

Factor . . .

Numerator: .$\displaystyle 4(x^2-16) \:=\:4(x-4)(x+4)$

Denominator: .$\displaystyle 12(x^2-5x+4) \:=\:12(x-1)(x-4)$

The fraction becomes: .$\displaystyle \frac{4(x-4)(x+4)}{12(x-1)(x-4)}$

Reduce: . $\displaystyle \frac{{\color{green}\rlap{/}}4\,{{\color{red}\rlap{//////}}(x-4)}\,(x+4)}{{\color{green}\rlap{//}}12_3\,(x-1)\,{\color{red}\rlap{//////}}(x-4)} \;=\;\boxed{\frac{x+4}{3(x-1)}}$

(2xsquared - 128) / (4xsquared - 36x + 32)
The same routine . . .

$\displaystyle \frac{2x^2-128}{4x^2-36x + 32} \;=\; \frac{\text{factor}}{\text{factor}} \;=\;\text{Reduce}$

4. (2xsquared - 128)/(4xsquared - 36x + 32)

2(xsquared - 64)
2(xsquared - 8squared)
2(x - 8)(x + 8)

4xsquared - 36x + 32
4(xsquared - 9x +8)
4(x - 1)(x - 8)

So, (2(x - 8)(x + 8))/(4x(x - 1)(x - 8))
= ((x + 8))/(2(x - 1))

Is this right?

EDIT: Soz, I forgot the parentheses

5. Originally Posted by Joker37
(2xsquared - 128)/(4xsquared - 36x + 32)

2(xsquared - 64)
2(xsquared - 8squared)
2(x - 8)(x + 8)

4xsquared - 36x + 32
4(xsquared - 9x +8)
4(x - 1)(x - 8)

So, (2(x - 8)(x + 8))/(4x(x - 1)(x - 8))
= ((x + 8))/(2(x - 1))

Is this right?

EDIT: Soz, I forgot the parentheses
You are correct:

$\displaystyle \frac{x+8}{2(x-1)}$