the problem is to simplify
(9x^2-12x / x^2+1) * (2x^2+2 / 9x^2+6x-24)
the answer is 2x / x+2
i just have no clue how to reach that conclusion...im really close though...im stuck at
2x(3x-4) / 3x^2+2x-8
any help would be great
thanks
Hello
Just before getting this result, you had something which looked like $\displaystyle \frac{6x(3x-4)}{9x^2+6x-24}$ (otherwise multiply the fraction by $\displaystyle \frac{3}{3}$ )
One can notice that the denominator is the beginning of the development of a square like $\displaystyle (a+b)^2=a^2+2ab+b^2$ : $\displaystyle 9x^2+6x-24=(3x)^2+2\cdot3\cdot x-24$. You should try to complete it and see what happens.
ah i totally forgot how to factor quadrilaterals with bigger numbers in front of the x2
the way i did it was to take $\displaystyle 9x^2+6x-24$ break it to $\displaystyle 3(3x^2+2x-8)$
then i did $\displaystyle 3*-8 = -24$ and found the two factors of -24 that equal 2, 6 and -4
so i had $\displaystyle (3x^2 +6x) (-4x -8) $ which factors to $\displaystyle 3x(x+2)-4(x+2) = (3x-4)(x+2)$
add the 3 in front of it again from the first factor and i got my answer
this is how i remember being taught, but is there a faster/more efficient way to come up with this factor?
and thank you all for the quick help
PS: flyingsquirrel & moo - Im not sure I really understand either of your posts :/ Moo's being the more confusing of the two
$\displaystyle (a+b)^2=a^2+2ab+b^2$ (1)
when you look at $\displaystyle \underbrace{9x^2}_{a^2}+\underbrace{2\cdot 3x \cdot1}_{2ab}-24$, you may notice that with $\displaystyle a=3x$ and $\displaystyle b=1$, the beginning of the expression looks like the beginning of (1).
The idea is to complete the square, that's to say to write $\displaystyle 9x^2+6x-24=\underbrace{9x^2}_{a^2}+\underbrace{2\cdot 3x\cdot1}_{2ab}+\underbrace{1}_{b^2}-24-1$.
It gives us $\displaystyle 9x^2+6x-24=(3x+1)^2-25=(3x+1)^2-5^2=(3x+1+5)(3x+1-5)$ which brings the simplifications you were looking for.
Using Moo's advice : finding the root(s) of the polynomial and then factoring it. (if $\displaystyle x_1$ and $\displaystyle x_2$ are the two roots of the polynomial, ($\displaystyle x_1=x_2$ allowed) $\displaystyle 9x^2+6x-24=9(x-x_1)(x-x_2)$)this is how i remember being taught, but is there a faster/more efficient way to come up with this factor?
I'm sorry about itPS: flyingsquirrel & moo - Im not sure I really understand either of your posts :/ Moo's being the more confusing of the two
It's a method you can understand only if you've already studied it
If you are interested : Solve Quadratic Equations Using Discriminants (1)
Good luck
I see how Moo's method would help me find the roots but I still don't get how after i have the roots i would come up with the factor of the polynomial.
Granted on this problem I dont think I could use that method anyway, considering the test im studying for does not allow calculators and the roots arent the easier numbers to deal with.
If $\displaystyle x_1$ and $\displaystyle x_2$ are the roots of $\displaystyle ax^2+bx+c$, we have the equality : $\displaystyle ax^2+bx+c=a(x-x_1)(x-x_2)$
Here, the roots were "easy" to calculateGranted on this problem I dont think I could use that method anyway, considering the test im studying for does not allow calculators and the roots arent the easier numbers to deal with.
$\displaystyle x_1=\frac{-2+6}{6}=\frac 23$
$\displaystyle x_2=\frac{-2-6}{6}=\frac{-4}{3}$
Hence $\displaystyle 3x^2+2x-8=3(x+\frac 43)(x-\frac 23)$
But don't use it if you haven't studied it yet, it'd be dangerous for your test