TheXSquaredFactor

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UsernameTheXSquaredFactor
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 #5
avatar+1602 
+1

 

Hello again! I see that you are having difficulty finishing the complete simplification of \(\frac{9x+2}{3x^2-2x-8}+\frac{7}{3x^2+x-4}\)

 

I am glad to see that you have factored both quadratics correctly. Rewriting the denominators into its factored form results in \(\frac{9x+2}{(3x+4)(x-2)}+\frac{7}{(3x+4)(x-1)}\). Just like the one I solved over here at https://web2.0calc.com/questions/subtracting-rational-expressions#r1, your goal at the moment is to create a common denominator. Unlike the one I linked to above, finding it is a little bit more difficult. In that example, I noted how you want the LCM (least common multiple); however, you really only need a common multiple. It does not have to be the least, but the computation is easier if you do. 

 

So, what is a common factor of \((3x+4)(x-2)\) and \((3x+4)(x-1)\)? The common factor of 3x+4 is relevant, but it alone will not be enough to figure out a common factor. To figure out a common multiple, just multiply both denominators together. This will always give you a common multiple. It may not be the least one, however. Multiplying both the denominators together yields \((3x+4)^2(x-2)(x-1)\). There is no need to expand yet. Notice, however, that both denominators have a factor of \(3x+4\), so we can divide that from the multiple we have. This gives us the least common multiple of \((x-2)(x-1)\).

 

Now, let's convert each fraction individually. I'll start with \(\frac{9x+2}{(3x+4)(x-2)}\).

 

\(\frac{9x+2}{(3x+4)(x-2)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\)I'm not necessarily concerned about expanding just yet. For now, let's just worry about getting the common denominator.
\(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}\) 
  

 

Now, let's convert the next fraction \(\frac{7}{(3x+4)(x-1)}\):

 

\(\frac{7}{(3x+4)(x-1)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\)Let's do the multiplication!
\(\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2}\) 
  

 

You might notice that there is an issue, though, because the expression \(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}+\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2} \Rightarrow \frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\). You'll see that I factored out some unnecessary factors to create that common denominator. Now, let's add those fractions together!

 

\(\frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\)Now that the denominators are the same, we can combine the fractions.
\(\frac{(9x+2)(x-1)+7(x-2)}{(3x+4)(x-2)(x-1)}\)

Now, let's multiply the numerator. Find the product of the binomials and distribute the 7.

\(\frac{9x^2-7x-2+7x-14}{(3x+4)(x-2)(x-1)}\)Let's combine those like terms!
\(\frac{9x^2-16}{(3x+4)(x-2)(x-1)}\)This is NOT the final answer. This is not fully simplified because the numerator is a difference of squares. Be attentive!
\(\frac{(3x+4)(3x-4)}{(3x+4)(x-2)(x-1)}\)There is a common factor in the numerator and denominator that cancel out.
\(\frac{3x-4}{(x-2)(x-1)}\)Now, expand the denominator again.
\(\frac{3x-4}{x^2-3x+2}\)This is fully simplified; there is nothing else to do here.
  

 

Now, let's consider the restrictions. All the factors of the original problem \((3x+4),(x-2),\text{and}(x-1)\) cannot be divided by zero. Set them equal to zero to figure out the forbidden values. 

 

\(\frac{3x-4}{x^2-3x+2},x\neq -\frac{4}{3},1,2\)

TheXSquaredFactor Jan 11, 2018
 #1
avatar+1602 
+1

In order to simplify this expression, we must create a common denominator. Otherwise, subtracting complicated fractions is not feasible. Currently, \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) does not have this, so we must create one. 

 

The denominator of the first term, \(a^2-1\) can be factored because it is a difference of two squares. Sometimes, factoring the denominator fully is helpful since it can make identifying the least common multiple all that more painless. In this case, \(a^2-1=(a+1)(a-1)\).

 

We now have \(\frac{a^2+1}{(a+1)(a-1)}-\frac{a-1}{a+1}\). Well, would you look at that! It is clear here that \((a+1)(a-1)\) is the least common multiple of the denominators since \(a+1\) is a factor of the product of binomials. Let's manipulate \(\frac{a-1}{a+1}\) so that we can create the desired common denominator.

 

\(\frac{a-1}{a+1}\) As aforementioned, let's convert this into a fraction with a common denominator.
\(\frac{a-1}{a+1}*\frac{a-1}{a-1}\) Of course, the actual value of the fraction remains unchanged since I am really multiplying the fraction by one. Now, let's combine.
\(\frac{(a-1)(a-1)}{(a+1)(a-1)}\)  
   

 

Ok, we have now changed the problem from \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) to \(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\). This transformation is indeed a positive one because we can now combine the fractions. We have now overcome the first stumbling block. 

 

\(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\) Combine the fractions now that the same denominator of both terms exist.
\(\frac{a^2+1-(a-1)(a-1)}{(a+1)(a-1)}\) Let's determine the product of the binomials. Since both binomials are identical, we can utilize the rule that \((x-y)^2=x^2-2xy+y^2\)
\(\frac{a^2+1-(a^2-2a+1)}{(a+1)(a-1)}\) Distribute the negative sign to all the terms it contains in parentheses.
\(\frac{a^2+1-a^2+2a-1}{(a+1)(a-1)}\) Look at that! In the numerator, the a2-term will cancel out and so will the constants!
\(\frac{2a}{(a+1)(a-1)}\) We can do the multiplication of the binomials in the denominator again. We already know what the product is.
\(\frac{2a}{a^2-1}\) Of course, let's not forget the restrictions for the value of a. 
\(\frac{2a}{a^2-1}, a\neq\pm1\) If a was equal to the restricted values, a division by zero would occur, which is forbidden.
TheXSquaredFactor Jan 11, 2018
 #1
avatar+1602 
+2

If the roots are \(x=2-\sqrt{3}\) and \(x=1-\sqrt{3}\), then the conjugates must also be roots as well, in order for it to be a polynomial. Therefore, \(x=2+\sqrt{3}\) and \(x=1+\sqrt{3}\) are also roots of this unknown polynomial.

 

With these roots, we can generate factors of the given polynomial, too.

 

\(x=2-\sqrt{3}\\ \hspace{1mm}-\left(2-\sqrt{3}\right)\) \(x=2+\sqrt{3}\\ \hspace{1mm}-\left(2+\sqrt{3}\right)\) \(x=1-\sqrt{3}\\ \hspace{1mm}-\left(1-\sqrt{3}\right)\) \(x=1+\sqrt{3}\\ \hspace{1mm}-\left(1+\sqrt{3}\right)\) By subtracting the right hand side of the equation, we can then see what the factors are.
\(x-\left(2-\sqrt{3}\right)=0\) \(x-\left(2+\sqrt{3}\right)=0\) \(x-\left(1-\sqrt{3}\right)=0\) \(x-\left(1+\sqrt{3}\right)=0\)  
\(x-2+\sqrt{3}=0\) \(x-2-\sqrt{3}=0\) \(x-1+\sqrt{3}=0\) \(x-1-\sqrt{3}=0\) Since all of these are set equal to zero, these all must be factors of the original unknown polynomial.
         

 

As aforementioned, they are factors of the original equation. If they are factors, then we can multiply them together and figure out what the original polynomial is.

 

\(\textcolor{red}{\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)}\textcolor{blue}{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}=0\)

 

We know that the left hand side is equal to zero since each factor equals zero. 0 multiplied by itself four times still equals zero. Now, expand. I recommend multiplying factors that are conjugates. This might make the process easier. We can multiply in any order, too. I'll expand the bit in red first.

 

\(\textcolor{red}{\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)}\) Make sure to remain attentive. Be sure to multiply every factor.
\(\textcolor{red}{x^2-2x-x\sqrt{3}-2x+4+2\sqrt{3}+x\sqrt{3}-2\sqrt{3}-3}\) Wow! Those are a lot of terms. Let's rearrange the equation, though, to see if any cancelling can occur here. (Spoiler: There is!)
\(\textcolor{red}{x^2-x\sqrt{3}+x\sqrt{3}+2\sqrt{3}-2\sqrt{3}-2x-2x+4-3}\) Look at that! All the radical expression cancel out! 
\(\textcolor{red}{x^2-4x+1}\) Wow, that monstrosity has been simplified quite beautifully. Don't you agree?
   

 

Let's do the exact same process with the blue bit. It should also simplify to a quadratic trinomial.

 

\(\textcolor{blue}{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}\) Ok, time to expand again! We already have an idea of what should occur.
\(\textcolor{blue}{x^2-x-x\sqrt{3}-x+1+\sqrt{3}+x\sqrt{3}-\sqrt{3}-3}\) Let's do that rearranging again!
\(\textcolor{blue}{x^2-x-x-x\sqrt{3}+x\sqrt{3}+\sqrt{3}-\sqrt{3}+1-3}\) Yet again, as expected, the radicals will cancel out.
\(\textcolor{blue}{x^2-2x-2}\)  
   

 

Now, we must multiply the red and blue parts together to get a quartic polynomial.

 

\((\textcolor{red}{x^2-4x+1})(\textcolor{blue}{x^2-2x-2})=0\) We have to expand one more time!
\(x^4-2x^3-2x^2-4x^3+8x^2+8x+x^2-2x-2=0\) Yet again, let's rearrange and see if any combining can take place.
\(x^4-2x^3-4x^3-2x^2+8x^2+x^2+8x-2x-2=0\) Ok, now let's combine.
\(x^4-6x^3+7x^2+6x-2=0\) We can turn this into a function.
   

 

\(f(x)=x^4-6x^3+7x^2+6x-2\)

 

Let's check to see that this function fits the strict conditions given in the original problem.

 

  • The function is monic because the leading coefficient is 1.
  • The function is quartic because the degree is 4.
  • The function is a polynomial because there is no division by a variable and no fractional exponents.
  • The function is written with \(f(x)=\).
  • All coefficients are rational.
  • The function has the given roots 
  • The answer is fully expanded and simplified

This polynomial fits these conditions, so this is the above is the answer.