Sup Veeky Forums i have this really basic question in my exam guide...

Sup Veeky Forums i have this really basic question in my exam guide, it should be really easy but o think the teacher wrote it wrong but maybe im just wrong so try to solve it guys ill post the answer in 15 minutes pld post the procedure btw

[eqn]
3 \left( 24x^{2} + 26x - 15 \right)
[/eqn]

I arrived there too and got it wrong btw you have to solve for x

...

ok I think you guys confirmed the exercise is wrong, the final answer is supposed to be 78

Second part

>solve for x
You only posted a shitty snapshot, so I typed it in wolfram alpha and pasted whatever it spat out.

ok I thought it was obvious but try it, hopefully I am wrong

solutions are: -1.5 and 0.4166667

1) Simplify everything up to : 72x^2+78x-45
2) Factor out the quadratics
3) (6x+9)(12x-5)=0

6x+9=0 , 12x-5=0
6x=-9 , 12x = 5
x=-9/6 , x=5/12

4) Simplify answer to : -3/2 and 5/12
Winrar

Ok try to simplify this now, sorry for the shitty quality

(x+4)(x-1) / (x+1)(x-1)

oops I mistyped

[math](x+4)(x+1)/(x+1)(1-x)[/math]

[eqn]
\begin{aligned}
3 \left( 24x^2 + 26x - 15 \right) &= 0 \\
3 \left( (2x+3)(12x+5) \right) &= 0 \\
x &= \left \{ -\frac{3}{2} , -\frac{5}{12} \right \}
\end{aligned}
[/eqn]
pls fug off now

[math] \frac{-4x^{4}+13x^{2}-9}{2x^{3}+3x^{2}-17x+12} [/math]
Did in a hurry, not sure.

I may be new to this board, but I could understand the sticky.

[eqn]
\frac{-(2x+3)(x+1)}{x+4}
[/eqn]

wrong
right pls post your procedure

I was getting lazy with LaTeX resizing and align (hence why I only posted the solution) but whatever I'll show the work.
[eqn]
\begin{align}
\frac{4x^2 - 9}{x^2 + 3x - 4} \cdot \frac{1 - x^2 }{2x - 3} &= \frac{(2x+3)(2x-3)}{(x-1)(x+4)} \cdot \frac{(1+x)(1-x)}{2x - 3} \\
&= \frac{(2x+3)(2x-3)}{(x-1)(x+4)} \cdot \frac{-(x+1)(x-1)}{2x - 3} \\
&= \frac{-(2x+3)(x+1)}{x+4}
\end{align}
[/eqn]

I would literally suck your cock thanks senpai

why did you add a - before (x+1)(x-1)

he factored out a negative from (1-x)

[math](1-x^2) = (1-x)(1+x) = -(-1+x)(1+x) = -(x-1)(x+1)[/math]

[math](1-x)^2[/math] should be the first one, sorry

You can't just add =0 to the end

What said. If I didn't factor out the negative, I wouldn't be able to cancel as much terms. Just remember to try that out whenever you have stuff like [math](a+x)(a - x)[/math].