Question

(4x+8)/(2x)
-
(x)/(x-4)
=
Once you find the answer discuss why the degree of the resulting denominator did not change from your expression’s degree.
PLEASE PLEASE PLEASE HELP!!!!!!!!!!!

Answer 1

Well the way you solve this is to factor the numerator. Like this:

Answer 2

When you do this you will get 4(x + 2)/2x. Looking at it more clearly,

Answer 3

\[\frac{ 4x+8 }{ 2x }-\frac{ x }{ x-4 }=\frac{ (4x+8)(x-4)-x(2x) }{ 2x(x-4) }\]

Answer 4

You can cancel between the 2 and the 4; the 2 from the 2x, not the one inside the parenthesis. That leaves an x only in the denominator. That is why the denominator's resulting degree did not change. You CANNOT cancel what is inside the parenthesis. You can' touch it! Only what is outside is legal!

Answer 5

Oh my...the problem is not just what I typed...it goes on and on...I didn't see that part as did neer2890! Oops!

Answer 6

\[=\frac{ 4x ^{2}-16x+8x-32-2x ^{2} }{ 2x(x-4) }=\frac{ 4x ^{2}-2x ^{2}-16x+8x-32 }{ 2x(x-4) }\]

Answer 7

So @neer2890 is correct?

Answer 8

Is that the final simplified answer

Answer 9

sorry.. i think i got a mistake
let me correct it.

Answer 10

ok

Answer 11

\[=\frac{ 2x ^{2}-8x-32 }{ 2x(x-4) }=\frac{ 2(x ^{2}-4x-16 )}{ 2x(x-4) }=\frac{ x ^{2}-4x-16 }{ x(x-4) }\]

Answer 12

Okay Thank you

Answer 13

you're welcome..:)
do you know the 2nd part?

Answer 14

No i don't. Would you be able to help me? :)

Answer 15

ok. you can see that the numerator of the fraction is a quadratic function
of the form\[ax ^{2}+bx+c\]
where a=1 ,b=-4 and c=-16

Answer 16

Yes i see

Answer 17

do you know how to factorize the quadratic functions?

Answer 18

Um...not really

Answer 19

to factorize quadratic functions of the form
\[ax ^{2}+bx+c\]
we first multiply first and 3rd term which gives us \[acx ^{2}\]
and now we have to factor this \[acx ^{2}\] into two parts such that it their addition becomes bx and their multiplication becomes \[acx ^{2}\]

Answer 20

Are you talking about how the resulting denominator did not change from the expression degree? the to figure out the answer to that question

Answer 21

yes

Answer 22

What exactly is that question asking of me? i really dont understand at all

Answer 23

ok. let me complete then you will understand what's going on.

Answer 24

okay

Answer 25

for e.g.,
\[x ^{2}-4x-32\]

first multiply x^2 and -32 which gives us\[-32x ^{2}\]
now we have to factor this \[-32x ^{2}\]
into two parts such that their addition becomes -4x and their multiplication becomes
\[-32x ^{2}\]

Answer 26

so +4x and -8x are the factors of this function.
and we can write it as
\[x ^{2}-4x-32\]
=\[x ^{2}+4x-8x-32\]
=x(x+4)-8(x+4)
=(x+4)(x-8)

Answer 27

this is called factorization of a quadratic polynomial
but you can see that we can not factorize
\[x ^{2}-4x-16\]

Answer 28

So theresulting denominator did not change from your expressions degree because you cannot factorize?

Answer 29

???

Answer 30

if we can't factorize the numerator then how come the denominator changes.
because if we can factor the numerator and one of the factor is (x-4) then it cancels out and the denominator changes.
But here, we can't factor the numerator that's why the denominator remains same.