# Thread: need help with doubt about exercise of polynomials (algebraic expressions)

1. ## need help with doubt about exercise of polynomials (algebraic expressions)

Hello to all the community , I'm new to this forum

I apologize if Iwrite something that is not understood , I will be very concrete.

The exercise is as follows.

calculate and indicate that values of X is defined the ratio.

My resolution

First thing I did was multiply polynomials crossed (x-3 ) * (8x + 16x^2)

and ( 4x ) * ( x
^2 - 9 )

and then re-divide the remaining fraction of polynomials but I realized that Ido not get anything.

Then I can not understand the part where it says

"for values of X is defined the ratio"

Please if anyone can help I would appreciate it !

Best of luck to everyone!

2. The bit about "indicate that values of X is defined the ratio" seems like it was poorly translated from another language. I don't understand it any better than you do, so I'd recommend asking your teacher for clarification, as only he/she can tell you what the worksheet is really asking. Until I know more about the problem, I don't feel comfortable conjecturing about what it might mean.

3. Originally Posted by au6us7o
I can not understand the part where it says

"for values of X is defined the ratio"
The directions are saying that, assuming the two sides are not always equal, find x-values for which the two sides are equal. For instance, in the following ratio:

. . . . .$\dfrac{x}{3}\, :\, \dfrac{2}{1}$

This is true exactly and only for:

. . . . .$1\, \cdot\, x\, =\, 2\, \cdot\, 3$

. . . . .$x\, =\, 6$

However, in the following case:

. . . . .$\dfrac{x}{3}\, :\, \dfrac{9x}{27}$

...the ratio is true for all values of the variable:

. . . . .$27x\, =\, 3(9x)$

. . . . .$27x = 27x$

This is trivially true if the variable equals zero, so assume not, and divide through:

. . . . .$27\, =\, 27$

So the value of the variable in this particular case is irrelevant; the ratio is true "for all values of" the variable.

. . . . .$\dfrac{x\, -\, 3}{4x}\, :\, \dfrac{x^2\, -\, 9}{8x\, +\, 16x^2}$

. . . . .$\dfrac{x\, -\, 3}{4x}\, :\, \dfrac{(x\, -\, 3)(x\, +\, 3)}{(4x)(2\, +\, 4x)}$

So what must be proportional to 1? Doing the cross-multiplication and solving the resulting equation, what do you get?

Originally Posted by au6us7o
calculate and indicate that values of X is defined the ratio.

$\mbox{a) }\, \dfrac{x\, -\, 3}{4x}\, :\, \dfrac{x^2\, -\, 9}{8x\, +\, 16x^2}$

First thing I did was multiply polynomials crossed:

$(x\, -\, 3)\, (8x\, +\, 16x^2)\, =\, (4x)\, (x^2\, -\, 9)$

and then re-divide the remaining fraction of polynomials but I realized that I do not get anything.
What did you do when you "re-divided"? What is the "remaining fraction of polynomials"? What did you get, that you concluded was "not ... anything"?

4. ## simplify?

I believe the problem wants you to simplify (reduce) the proportion.
Perhaps:

1 : (x+3)/(4x+2) when (x-3)#0 and x#0

or

(4x+2) : (x+3) when x#-1/2, x#3 and x#0