Question

-Using complete sentences, explain how the Rational Root Theorem and Descartes' Rule of Signs are used to find zeros of a polynomial function.
-Simplify.
[(x^2+2x)/(12x+54)]-[(3-x)/(8x-36)]
Someone please help me!!

Answer 1

do v need to solve for x?

Answer 2

\[\LARGE \frac{(x^2+2x)}{(12x+54)}-\frac{(3-x)}{(8x-36)}\]
this looks better... @nahomie you have to simplify right? ... i think we can't solve for "x" since there's no equal sign ! O_O

Answer 3

yeah, just simplifying. i'm not looking for x. and it does look better, how do i pt in my questions to look like that?

Answer 4

pretty difficult task for simple expression.

Answer 5

\[\Large \frac{(x^2+2x)\cdot(8x-36)}{(12x+54)(8x-36)}-\frac{(12x+54)(3-x)}{(12x+54)(8x-36)}\]
\[\large \frac{(8x^3-36x^2+16x^2-72x)}{(96x^2-432x+432x-1944 )}-\frac{( 36x-12x^2+162-54x)}{(96x^2-432x+432x-1944 )}\]
\[\large \frac{(8x^3-36x^2+16x^2-72x)-( 36x-12x^2+162-54x)}{(96x^2-432x+432x-1944 )}\]
\[\large \frac{(8x^3-36x^2+16x^2-72x-36x+12x^2-162+54x)}{(96x^2-1944 )}\]
\[\large \frac{(8x^3-4x^2-54x-162)}{(96x^2-1944 )}\]

Answer 6

this wasn't easy at all! :(

Answer 7

thnk u everybody, i appreciate it