Rational expression

**silencecloak** · May 4th 2008, 01:33 AM

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

**flyingsquirrel** · May 4th 2008, 01:51 AM

Hello

Just before getting this result, you had something which looked like $\frac{6x(3x-4)}{9x^2+6x-24}$ (otherwise multiply the fraction by $\frac{3}{3}$

)

One can notice that the denominator is the beginning of the development of a square like $(a+b)^2=a^2+2ab+b^2$ : $9x^2+6x-24=(3x)^2+2\cdot3\cdot x-24$ . You should try to complete it and see what happens.

**Moo** · May 4th 2008, 01:53 AM

Originally Posted by silencecloak

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,

Factor 3x²+2x-8.

$\Delta=4+4\cdot8=6^2$

--> $x_1=\frac{-2+6}{6}$ and $x_2=\frac{-2-6}{6}$

$3x^2+2x-8=3(x-x_1)(x-x_2)$

**wingless** · May 4th 2008, 01:54 AM

$\frac{9x^2-12x}{x^2 +1}\cdot\frac{2x^2+2}{9x^2+6x-24}$

Factorise all.

$\frac{3x(3x-4)}{x^2 +1}\cdot\frac{2(x^2+1)}{3(x+2)(3x-4)}$

**silencecloak** · May 4th 2008, 02:14 AM

ah i totally forgot how to factor quadrilaterals with bigger numbers in front of the x2

the way i did it was to take $9x^2+6x-24$ break it to $3(3x^2+2x-8)$

then i did $3*-8 = -24$ and found the two factors of -24 that equal 2, 6 and -4

so i had $(3x^2 +6x) (-4x -8)$ which factors to $3x(x+2)-4(x+2) = (3x-4)(x+2)$

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

**flyingsquirrel** · May 4th 2008, 02:53 AM

Originally Posted by silencecloak

PS: flyingsquirrel & moo - Im not sure I really understand either of your posts :/ Moo's being the more confusing of the two

$(a+b)^2=a^2+2ab+b^2$ (1)
when you look at $\underbrace{9x^2}_{a^2}+\underbrace{2\cdot 3x \cdot1}_{2ab}-24$ , you may notice that with $a=3x$ and $b=1$ , the beginning of the expression looks like the beginning of (1).
The idea is to complete the square, that's to say to write $9x^2+6x-24=\underbrace{9x^2}_{a^2}+\underbrace{2\cdot 3x\cdot1}_{2ab}+\underbrace{1}_{b^2}-24-1$ .
It gives us $9x^2+6x-24=(3x+1)^2-25=(3x+1)^2-5^2=(3x+1+5)(3x+1-5)$ which brings the simplifications you were looking for.

this is how i remember being taught, but is there a faster/more efficient way to come up with this factor?

Using Moo's advice : finding the root(s) of the polynomial and then factoring it. (if $x_1$ and $x_2$ are the two roots of the polynomial, ( $x_1=x_2$ allowed) $9x^2+6x-24=9(x-x_1)(x-x_2)$ )

**Moo** · May 4th 2008, 03:00 AM

PS: flyingsquirrel & moo - Im not sure I really understand either of your posts :/ Moo's being the more confusing of the two

I'm sorry about it

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

**silencecloak** · May 4th 2008, 03:14 AM

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.

**Moo** · May 4th 2008, 03:24 AM

Originally Posted by silencecloak

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.

If $x_1$ and $x_2$ are the roots of $ax^2+bx+c$ , we have the equality : $ax^2+bx+c=a(x-x_1)(x-x_2)$

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.

Here, the roots were "easy" to calculate

$x_1=\frac{-2+6}{6}=\frac 23$

$x_2=\frac{-2-6}{6}=\frac{-4}{3}$

Hence $3x^2+2x-8=3(x+\frac 43)(x-\frac 23)$

But don't use it if you haven't studied it yet, it'd be dangerous for your test

**silencecloak** · May 4th 2008, 11:20 AM

Originally Posted by Moo

If $x_1$ and $x_2$ are the roots of $ax^2+bx+c$ , we have the equality : $ax^2+bx+c=a(x-x_1)(x-x_2)$

Here, the roots were "easy" to calculate

$x_1=\frac{-2+6}{6}=\frac 23$

$x_2=\frac{-2-6}{6}=\frac{-4}{3}$

Hence $3x^2+2x-8=3(x+\frac 43)(x-\frac 23)$

But don't use it if you haven't studied it yet, it'd be dangerous for your test

Oh, well ya I see that but I got lost on how to take that to the next step so I can cross cancel other factors with it.

Thank you for taking the time to explain it to me though.

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