Question

Help in simplifying rational expressions. MEDAL!!!

Answer 1

\[\large \frac{ 2 }{ x-4 } - \frac{ 2x }{ 7x - 8 }\]

Answer 2

@Hero @pooja195

Answer 3

@jigglypuff314

Answer 4

I had the fraction (which still needs to be simplified)
\[\large \frac{ 14x - 16 - 2x^2 + 8x }{ 7x^2 - 36x + 32 } = \frac{ -2x^2 - 6x - 16 }{ 7x^2 - 36x + 32 }\]

Answer 5

14x + 8x ?

Answer 6

When I try to factor \(\large -2x^2 - 6x - 16\) I would have \(\large 2(-x^2 - 3x - 8)\) but when I try to simplify that I don't get two factors of -8 when added together will give me -3.

Answer 7

Oh okay!

Answer 8

I now have:
\(2(-x^2 + 11x - 8)\)
is that correct for factoring?

Answer 9

@jigglypuff314

Answer 10

that's what I got too...
but I'm not sure how to factor it, sorry

Answer 11

yes thats correct :)
so u factored the numerator
now tell this-> what will be in the denominator

Answer 12

why can't i factor it anymore?

Answer 13

you can factor it but the factors will look horrible

if you have any equation like this-> \(\large ax^2 +bx +c\)
we know that this is a quadratic equation and it can be written like this->\(a(x-\alpha)(x-\beta)\)
where \(\alpha~and~\beta\) are the roots of the polynomial

Now the roots are given by this formula->\(\Large {\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

so one root say \(\alpha\) is this->\(\Large {\frac{-b + \sqrt{b^2-4ac}}{2a}}\)

and the other root \(\beta\) is this->\(\Large {\frac{-b - \sqrt{b^2-4ac}}{2a}}\)

now just put the value of \(\alpha\) and \(\beta\) in our factored form->\(a(x-\alpha)(x-\beta)\)
then our polynomial=factored :)


note that this is only for quadratic polunomials