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$${3x + 15 \over x^{2} - 25} +{4x^{2} - 1 \over 2x^{2} + 9x - 5}$$

the answer is $${2(x^{2} - 3x + 5) \over (x - 5)(x + 5)} , x ≠±5, {1 \over 2}$$

here's what i did: but i didn't get the answer, what did i do wrong?

Jul 23, 2018

#1
+1

Hello Guest,

I have not scrutinized your work thoroughly yet, but I think you have done everything correctly thus far. In order to make progress here, you must simplify the numerator $$3(2x^2+9x-5)+(4x^2-1)(x-5)$$. Do you know how to proceed from here?

Jul 24, 2018
edited by TheXSquaredFactor  Jul 24, 2018
#2
+1

I multiply 3 with the trinomial, then expand the two other binomials?

Guest Jul 24, 2018
#3
+1

Absolutely! Otherwise, the expression is not completely simplified.

TheXSquaredFactor  Jul 24, 2018
#4
+1

i got $${4x^{3} - 14x^{2} + 17x -10 \over (x +5)(x+5)(2x-1)}$$, and i'm not sure on what to do next?

Guest Jul 24, 2018
#5
+1

Hello Guest,

You have been proactive in attempting to simplify the numerator, and you are continuing to make significant headway. However, there is a slight algebraic error lying somewhere in your simplification, but I cannot pinpoint where because of the lack of work provided. Computations like these require extreme perserverance and accuracy. Any small imperfection will result in an answer that is horribly askew.

 $$3(2x^2+9x-5)+(4x^2-1)(x-5)$$ For now, I am worrying about the simplification of the numerator. Let's distribute the 3 into the trinomial. $$6x^2+27x-15+(4x^2-1)(x-5)$$ It is time to expand the product of two binomials. $$\textcolor{red}{6x^2}\textcolor{blue}{+27x}\textcolor{green}{-15}+4x^3\textcolor{red}{-20x^2}\textcolor{blue}{-x}\textcolor{green}{+5}$$ Combine all like terms. I have utilized color coding to show which terms combine. $$4x^3-14x^2+26x-10$$

After this algebra, we are left with $$\frac{4x^3-14x^2+26x-10}{(x+5)(x-5)(2x-1)}$$. Notice the discrepancies and try and locate the mistake.

From here, you must factor the numerator and determine whether or not any of the factors of $$4x^3-14x^2+26x-10$$ match those of the denominator. Factoring by grouping does not appear to be a valid option here, so you will most likely have to resort to methods such as the rational root theorem and the Descartes' Rules of Signs. To start, though, notice how each term is even, so one can factor out a 2 from every term.

Good luck!

TheXSquaredFactor  Jul 24, 2018
#6
0

You make it unnecessarily complicated, it's usually a good idea to tidy up as you go along.

Look at your fourth line down, $$\displaystyle \frac{3(x+5)}{(x-5)(x+5)}+\frac{(2x-1)(2x+1)}{(x+5)(2x-1)} \; .$$

Why not cancel the (x + 5)'s in the first fraction and the (2x - 1)'s in the second  ?

That gets you $$\displaystyle \frac{3}{(x-5)}+\frac{(2x+1)}{(x+5)}\; ,$$

which is much easier to handle.

Jul 24, 2018
#7
+1

Yes, you are right. This route is much better.

TheXSquaredFactor  Jul 24, 2018